3.1 Inequalities and absolute value

Motivation

An inequality is an assertion about order, so its algebra has a direction. Adding the same real number to both sides preserves that direction, whereas multiplication by a negative number reverses it. Multiplication by an expression whose sign is unknown is therefore not a harmless simplification: it is a request to split the argument into cases. The same discipline governs squaring, taking reciprocals, and clearing denominators.

Inequalities serve three related purposes in this course. First, they describe solution sets, often as unions or intersections of intervals. Second, they prove comparisons such as AM-GM and Cauchy-Schwarz, with equality cases that reveal when a bound is sharp. Third, they provide estimates. An exact value may be difficult to compute, but a bound can control a limit, show that a polynomial eventually has a fixed sign, or quantify how close the terms of a sequence are to a proposed limit.

The reliable habit is always the same: record the domain, identify every possible sign change, justify each transformation, and intersect the result with the original domain. A plausible final interval is not a proof unless those obligations have been met.

Solving inequalities: domain before signs

The first task is to preserve the same solution set through each algebraic step. Read the sign of a multiplier before using it.

Definition

Solution set and critical points

The solution set of an inequality is the subset of its domain on which the statement is true. Zeros of the numerator, zeros of the denominator, and breakpoints of absolute values are critical points: between consecutive critical points, the relevant signs are constant. A denominator zero is never admitted, even when an algebraic cancellation seems to remove it.

Theorem

Order laws and safe transformations

For real $a,b,c$, trichotomy and transitivity hold. Moreover,

$$a\lt b\Longrightarrow a+c\lt b+c, \qquad \begin{cases} ac\lt bc,&c\gt0,\\ ac\gt bc,&c\lt0. \end{cases}$$

If $c=0$, multiplication destroys the comparison rather than preserving an equivalent inequality. If $0\lt a\lt b$ and $r\gt0$, then $a^r\lt b^r$ and $a^{-r}\gt b^{-r}$; the positivity assumption is essential for arbitrary real powers and reciprocals.

Checkpoint

Why is multiplying an inequality by $x-1$ not automatically reversible?

Identify all three possible signs of the multiplier.

Solution · Quick-check answer 1

For $x\gt1$ the direction is preserved, for $x\lt1$ it is reversed, and at $x=1$ the multiplier is zero; if it came from a denominator, that point is outside the domain.

Worked example

Three domain-first sign analyses

First solve a rational inequality. The domain excludes $x=1$, and moving to one side gives

$$\frac{x+1}{x-1}\le2 \Longleftrightarrow \frac{x-3}{x-1}\ge0 \Longleftrightarrow x\lt1\ \text{or}\ x\ge3.$$

One may instead multiply by $(x-1)^2\gt0$, but only after recording $x\ne1$.

Next, for $x\ne0$,

$$x\gt\frac3x+2 \Longleftrightarrow \frac{(x-3)(x+1)}{x}\gt0 \Longleftrightarrow -1\lt x\lt0\ \text{or}\ x\gt3.$$

Finally, $e^{-x}\sqrt{x-1}$ is positive exactly when $x\gt1$. On that domain,

$$\frac{(x-2)^3(x-3)}{e^{-x}\sqrt{x-1}}\gt0 \Longleftrightarrow 1\lt x\lt2\ \text{or}\ x\gt3.$$

The strict inequality excludes the numerator zeros $2$ and $3$.

Common mistake

Clearing a denominator of unknown sign

Multiplying by $x-1$ without cases can reverse the inequality or include the forbidden point $x=1$. Use sign cases, a sign chart, or the positive square $(x-1)^2$ after stating the exclusion.

Common mistake

Squaring before controlling signs

From $u\lt v$, one cannot conclude $u^2\lt v^2$ without suitable sign information. Likewise, arbitrary real powers require positive bases. Establish nonnegativity or split into cases before using a supposedly monotone operation.

Absolute value: distance and cases

A distance condition turns into intervals on the real line. Triangle inequalities then compare distances without requiring an exact value.

Definition

Absolute value as distance

For a real number $t$, absolute value is defined by

$$|t|:= \begin{cases} t,&t\ge 0,\\ -t,&t\lt0. \end{cases}$$

Thus $|t|$ is the distance from $t$ to $0$, while $|x-a|$ is the distance from $x$ to $a$. In particular, $|t|\ge 0$ and $|t|=\sqrt{t^2}$, where the square root is the nonnegative one.

Theorem

Absolute-value and triangle inequalities

For real $a,b$ and $r\ge0$,

$$|a|\le r\Longleftrightarrow -r\le a\le r, \qquad |a|\ge r\Longleftrightarrow a\le-r\ \text{or}\ a\ge r.$$

When $r\gt0$, the strict forms are

$$|a|\lt r\Longleftrightarrow -r\lt a\lt r, \qquad |a|\gt r\Longleftrightarrow a\lt-r\ \text{or}\ a\gt r.$$

Absolute value also satisfies

$$|-a|=|a|, \qquad |ab|=|a||b|, \qquad -|a|\le a\le |a|.$$

Moreover,

$$|a+b|\le |a|+|b|, \qquad \bigl||a|-|b|\bigr|\le |a-b|\le |a|+|b|.$$

Equality in $|a+b|\le|a|+|b|$ holds exactly when $ab\ge0$. For $n\in\mathbb Z_+$ and real $a_1,\ldots,a_n$, repeated application gives

$$|a_1+\cdots+a_n|\le |a_1|+\cdots+|a_n|.$$

Equality holds exactly when all summands are nonnegative or all are nonpositive.

Proof. The triangle inequality follows by comparing squares. Because both sides are nonnegative,

$$|a+b|^2=a^2+2ab+b^2 \le |a|^2+2|a||b|+|b|^2=(|a|+|b|)^2.$$

Replacing $a$ by $a-b$ yields one side of the reverse triangle inequality; swapping $a$ and $b$ supplies the other side.

Worked example

Piecewise absolute-value inequalities

For $|x-2|+|2x+1|\ge4$, assign each breakpoint to exactly one interval:

$$|x-2|+|2x+1|= \begin{cases} -3x+1,&x\lt-\tfrac12,\\ x+3,&-\tfrac12\le x\lt2,\\ 3x-1,&x\ge2. \end{cases}$$

Solving within each interval and then taking the union gives

$$|x-2|+|2x+1|\ge4 \Longleftrightarrow x\le-1\ \text{or}\ x\ge1.$$

Equivalently, graph the correctly labelled function $y=|x-2|+|2x+1|$ and compare it with the horizontal line $y=4$; the piecewise algebra explains exactly why the same two rays appear.

Two further case analyses give

$$|x-2|\lt x^2\Longleftrightarrow x\lt-2\ \text{or}\ x\gt1,$$

and, after excluding $x=-1$ and respecting the negative denominator,

$$\frac{|x+2|}{x+1}\lt-1 \Longleftrightarrow -\frac32\lt x\lt-1.$$

Common mistake

Losing the logic of absolute value

$|x|$ is not automatically $x$. A small-distance condition produces an intersection, whereas a large-distance condition normally produces a union. Breakpoints must be assigned consistently so that no point is omitted or counted under contradictory formulas.

Checkpoint

Rewrite $|x-5|\lt2$ as a single interval.

Interpret the expression as distance from $5$.

Solution · Quick-check answer 2

The condition is $-2\lt x-5\lt2$, hence $3\lt x\lt7$.

Positive sums and nonnegative squares

The next task is to prove a comparison that holds for every admissible input. Positive weights and nonnegative squares make both the direction and the equality case visible.

Theorem

A positive weighted ratio lies between its endpoints

Let $n\in\mathbb Z_+$, let $y_i\gt0$, and let $m\le x_i/y_i\le M$ for $1\le i\le n$. Then

$$m\le \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n y_i}\le M.$$

The left inequality is strict if at least one ratio is strictly larger than $m$; the right inequality is strict if at least one is strictly smaller than $M$. Hence, when $n\ge2$ and $x_1/y_1\lt\cdots\lt x_n/y_n$, the ratio of sums lies strictly between the two endpoint ratios. Taking $x_i=\sin\alpha_i$ and $y_i=\cos\alpha_i$ is legitimate for $0\lt\alpha_i\lt\pi/2$, because every denominator is positive.

As a basic special case, if $p\ge a\gt0$ and $b\ge q\gt0$, then $aq\le pq\le pb$, so division by $bq\gt0$ gives $a/b\le p/q$.

Proof. The ratio theorem is a useful model for strict endpoint reasoning. From $m\le x_i/y_i\le M$ and $y_i\gt0$, one gets $my_i\le x_i\le My_i$ term by term. After summation, division by the positive number $\sum_i y_i$ gives the weak bounds. If at least one lower comparison is strict, its positive gap survives the sum, so the lower bound is strict; the upper endpoint is handled separately in the same way. One must not pretend that every term is strict when an endpoint ratio is actually attained.

Worked example

Nonnegative squares and a strict weighted ratio

For $x,y\gt0$, division by $xy\gt0$ preserves order and gives

$$(x-y)^2\ge0 \Longrightarrow x^2+y^2\ge2xy \Longrightarrow \frac{x}{y}+\frac{y}{x}\ge2.$$

Equality holds exactly when $x=y$. Similarly, for real $a,b,c$,

$$(a-b)^2+(b-c)^2+(c-a)^2 =2(a^2+b^2+c^2-ab-bc-ca)\ge0,$$

so $a^2+b^2+c^2\ge ab+bc+ca$, with equality exactly when $a=b=c$.

For the ratio theorem, multiply the weak bounds $m\le x_i/y_i\le M$ by the positive number $y_i$ and sum. The lower bound becomes strict if at least one ratio exceeds $m$; separately, the upper bound becomes strict if at least one ratio is below $M$. For example, $x_1=1$, $x_2=2$, and $y_1=y_2=1$ attain the endpoint ratios $m=1$ and $M=2$, but their ratio of sums is $3/2$, strictly between them.

Means, sharp bounds, and equality

Means package several inputs into one representative value. Their inequalities are useful only together with their domains and equality conditions.

Definition

Four classical means

Let $n\in\mathbb Z_+$. For positive real numbers $a_1,\ldots,a_n$, define the arithmetic, geometric, harmonic, and quadratic means by

$$A_n=\frac1n\sum_{i=1}^n a_i, \qquad G_n=\left(\prod_{i=1}^n a_i\right)^{1/n}, \qquad H_n=\frac{n}{\sum_{i=1}^n 1/a_i}, \qquad Q_n=\left(\frac1n\sum_{i=1}^n a_i^2\right)^{1/2}.$$

$A_n$ and $Q_n$ make sense for arbitrary real inputs; $G_n$ in the AM-GM theorem allows nonnegative inputs; $H_n$ requires positive inputs.

Theorem

AM-GM and weighted AM-GM

For $n\in\mathbb Z_+$ and nonnegative $a_1,\ldots,a_n$,

$$\frac{a_1+\cdots+a_n}{n}\ge (a_1\cdots a_n)^{1/n},$$

with equality exactly when all inputs are equal. For $a,b\gt0$ and $0\lt p\lt1$,

$$a^p b^{1-p}\le pa+(1-p)b,$$

again with equality exactly when $a=b$. Equivalently, if $\alpha,\beta>0$, then

$$a^\alpha b^\beta\le \left(\frac{\alpha a+\beta b}{\alpha+\beta}\right)^{\alpha+\beta}.$$

Proof. For a concise proof of general AM-GM, first use calculus to obtain $\log t\le t-1$ for $t\gt0$. If every $a_i$ is positive and $A=(a_1+\cdots+a_n)/n$, then

$$\sum_{i=1}^n\log\frac{a_i}{A} \le \sum_{i=1}^n\left(\frac{a_i}{A}-1\right)=0.$$

Exponentiating gives $\prod_i a_i\le A^n$. Equality forces every $a_i/A=1$. If some input is zero, the geometric mean is zero; the arithmetic mean is nonnegative, and equality is possible only when every input is zero. This handles the zero case before using logarithms or cancelling a positive mean.

Weighted AM-GM is the two-term version with weights. The calculus inequality $t^p\le pt+1-p$ for $t\gt0$ and $0\lt p\lt1$, applied to $t=a/b$ and then multiplied by $b\gt0$, gives $a^pb^{1-p}\le pa+(1-p)b$. The equality condition $t=1$ becomes $a=b$.

Worked example

AM-GM applications and two Euler companions

For positive $a,b,c$, AM-GM applied to $a/b,b/c,c/a$ gives

$$\frac ab+\frac bc+\frac ca\ge3.$$

Applying AM-GM to the positive reciprocals gives $H_n\le G_n$. Applying it to $n$ copies of $1+1/n$ together with one copy of $1$ proves that $u_n=(1+1/n)^n$ is strictly increasing. A parallel application to suitable reciprocals proves that $v_n=(1+1/n)^{n+1}$ is strictly decreasing. Positivity is required throughout; the strictness comes from the entries not all being equal.

More explicitly, the first application has arithmetic mean $1+1/(n+1)$ and geometric mean $((1+1/n)^n)^{1/(n+1)}$, so raising a positive strict inequality to the power $n+1$ gives $u_{n+1}>u_n$. For the companion, apply AM-GM to $n+1$ copies of $n/(n+1)$ and one copy of $1$. Their arithmetic mean is $(n+1)/(n+2)$; raising the strict comparison to the power $n+2$ and then taking positive reciprocals yields $v_{n+1}\lt v_n$.

Weighted AM-GM also yields, for $u,v\gt0$,

$$u^{1/3}v^{2/3}\le\frac13u+\frac23v.$$

Theorem

Cauchy-Schwarz and the hierarchy of means

For $n\in\mathbb Z_+$ and real $x_i,y_i$,

$$\left(\sum_{i=1}^n x_i y_i\right)^2 \le \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right).$$

Equality holds exactly when the two vectors are linearly dependent, including the case in which one vector is zero. Only for two nonzero vectors may the normalized inner product be interpreted as a cosine. For arbitrary real $a_i$, Cauchy-Schwarz gives $Q_n\ge|A_n|$. For positive $a_i$, the full chain is

$$Q_n\ge A_n\ge G_n\ge H_n,$$

and equality throughout occurs exactly when all inputs are equal.

Proof. For Cauchy-Schwarz, dispose of the zero-vector case first. If $\sum_i x_i^2\gt0$, consider

$$F(t)=\sum_{i=1}^n(tx_i-y_i)^2 =\left(\sum x_i^2\right)t^2-2\left(\sum x_i y_i\right)t+\sum y_i^2.$$

Since $F(t)\ge0$ for every real $t$, its discriminant is nonpositive, which is exactly the stated inequality. Equality means $F(\lambda)=0$ for some $\lambda$, so every $y_i=\lambda x_i$.

Worked example

Cauchy-Schwarz in Engel form

When $b_i\gt0$, apply Cauchy-Schwarz to $x_i=a_i/\sqrt{b_i}$ and $y_i=\sqrt{b_i}$:

$$\sum_{i=1}^n\frac{a_i^2}{b_i} \ge\frac{(a_1+\cdots+a_n)^2}{b_1+\cdots+b_n}.$$

For positive $a,b,c$, choosing $x=(\sqrt{a^3b},\sqrt{b^3c},\sqrt{c^3a})$ and $y=(1/\sqrt{ab},1/\sqrt{bc},1/\sqrt{ca})$ gives

$$abc(a+b+c)\le a^3b+b^3c+c^3a.$$

The square roots and reciprocals explain why positivity, not mere reality, is part of the hypothesis in this application. Here $1/(ab)+1/(bc)+1/(ca)=(a+b+c)/(abc)$; dividing the Cauchy bound by the positive quantity $a+b+c$ completes the displayed conclusion.

Engel form also solves a cyclic example. If $x,y,z,w\gt0$ and $xyzw=16$, then

$$\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+w}+\frac{w^2}{w+x} \ge\frac{x+y+z+w}{2}\ge4.$$

The last step is AM-GM, and equality throughout requires $x=y=z=w=2$.

Common mistake

Using a named inequality outside its hypotheses

General AM-GM permits nonnegative inputs, but logarithmic proofs and weighted real powers require positive inputs. HM requires positive denominators; Engel form requires $b_i\gt0$; the cosine interpretation of Cauchy-Schwarz requires two nonzero vectors.

Checkpoint

For arbitrary real inputs, what stronger estimate involving $|A_n|$ can replace $Q_n\ge A_n$?

Apply Cauchy-Schwarz to $(a_1,\ldots,a_n)$ and $(1,\ldots,1)$.

Solution · Quick-check answer 3

Cauchy-Schwarz gives $Q_n\ge|A_n|\ge A_n$ for arbitrary real inputs. Thus $Q_n\ge A_n$ is already valid; the absolute-value bound is stronger. The full chain $Q_n\ge A_n\ge G_n\ge H_n$ is stated for positive inputs.

Turning bounds into control

An estimate becomes more powerful when its bound can be made as small or as large as needed. Here the same absolute-value tools control a local limit and the eventual sign of a polynomial.

Definition

A punctured limit

For $f:\mathbb R\setminus\{c\}\to\mathbb R$, the assertion $\lim_{x\to c}f(x)=L$ means

$$(\forall\varepsilon>0)(\exists\delta>0) (\forall x\in\mathbb R\setminus\{c\}) \bigl(0\lt|x-c|\lt\delta\Longrightarrow |f(x)-L|\lt\varepsilon\bigr).$$

The exclusion $x\ne c$ is encoded by $0\lt|x-c|$. The number $\delta$ may depend on $\varepsilon$, but not on the subsequently chosen $x$.

Worked example

Absolute-value estimates and epsilon control

The reverse triangle inequality avoids solving an entire compound inequality:

$$|2x-1|+|x|\le5 \Longrightarrow 2|x|-1+|x|\le5 \Longrightarrow |x|\le2.$$

For the cubic, let $\delta=\min\{\varepsilon/20,1\}$. If $|x-2|\lt\delta$, then $|x|\lt3$, and hence

$$|x^3-8| =|x-2||x^2+2x+4| \le |x-2|(|x|^2+2|x|+4) \lt20\delta\le\varepsilon.$$

The preliminary restriction $\delta\le1$ controls the otherwise variable factor. Any smaller positive $\delta$ also works, so the choice is not unique. For $f(x)=x^2+4x$, the related local estimate is

$$|x+3|\lt1\Longrightarrow |f(x)+3| =|x+3||x+1|\le3|x+3|.$$

Using a non-strict final bound makes the statement valid also at $x=-3$. If $0\lt\varepsilon\lt1$, the choice $\delta=\varepsilon/3$ then gives the required strict $\varepsilon$ estimate.

Theorem

Dominance of the leading polynomial term

Let $p(x)=ax^3+bx^2+cx+d$ with $a\gt0$. There are numbers $m\lt0\lt M$ such that $p(m)\lt0\lt p(M)$. Since a polynomial is continuous, the Intermediate Value Theorem then gives a real zero between $m$ and $M$. The same dominance argument applies to every odd-degree real polynomial whose leading coefficient is nonzero.

Worked example

Polynomial dominance and a quantitative sequence bound

Let $B=|b|+|c|+|d|$. For $x\ge1$,

$$p(x)=ax^3+bx^2+cx+d \ge ax^3-Bx^2=x^2(ax-B).$$

Thus, for any $\alpha\gt0$, choosing $M\ge\max\{1,(B+\alpha)/a\}$ ensures $p(x)\gt\alpha$ whenever $x\gt M$. For $R\ge1$, the corresponding estimate $p(-R)\le-aR^3+BR^2=R^2(B-aR)$ is negative once $R\gt B/a$. Continuity therefore gives the real root asserted above, without reusing the coefficient $c$ as the name of the root.

For $x_n=(n^2-3)/(n^2-5n-1)$ and $n\ge12$,

$$|x_n-1| =\left|\frac{5n-2}{n^2-5n-1}\right| \le\frac{7n}{n^2/2}=\frac{14}{n}.$$

Consequently, for $\varepsilon_0=2^{-1025}$, the explicit choice

$$N=14\cdot2^{1025}$$

satisfies $|x_n-1|\le\varepsilon_0$ for every integer $n\ge N$. The exponent in $\varepsilon_0$ remains negative; only its reciprocal appears in $N$.

A related calculus estimate

Calculus also proves $e^x\ge1+x$: the function $e^x-x-1$ decreases to $0$ on the negative half-line and increases from $0$ on the positive half-line. Thus $0$ is its global minimum, and equality occurs only at $x=0$.

Summary

Solving an inequality is a domain-and-sign argument. Translate all terms to one side, mark numerator zeros, denominator zeros, and absolute-value breakpoints, then test the constant-sign intervals. Squaring or clearing denominators is safe only after its sign conditions are explicit.

For proofs, nonnegative squares lead naturally to the two-variable AM-GM inequality, triangle inequality, and many elementary comparisons. General AM-GM, weighted AM-GM, and Cauchy-Schwarz package those ideas into reusable bounds. Equality conditions are part of each theorem, not optional decoration. Finally, triangle estimates turn local information into epsilon control, while leading-term estimates govern polynomials and sequences.

Exercises

1. For every integer $n\ge2$, prove by induction that $(1\cdot3\cdots(2n-1))/(2\cdot4\cdots2n)\lt1/\sqrt{3n+1}$.

2. For distinct positive $a,b$, prove $a^{n+1}-a^n b>ab^n-b^{n+1}$ for every positive integer $n$. Then prove $b^n((n+1)a-nb)\lt a^{n+1}$ and use it to show that $(1+1/n)^n$ is strictly increasing.

3. Prove Bernoulli's inequality $(1+x)^n\ge1+nx$ for $x\gt-1$ and positive integers $n$. State when equality occurs.

4. Solve $(x+1)/(x-1)\le2$ in two ways: by sign cases and by multiplication with a positive square. Explain why $x=1$ is excluded in both arguments.

5. Solve $|x-2|\lt x^2$ and $|x+2|/(x+1)\lt-1$, recording every breakpoint and forbidden value.

6. Use AM-GM to prove $a/b+b/c+c/a\ge3$ for positive $a,b,c$, and determine the equality case. Also prove $H_n\le G_n$ for positive inputs.

7. Prove Engel's form of Cauchy-Schwarz for real $a_i$ and positive $b_i$. Then apply it with $a_i=\sin\theta_i$ and $b_i=\cos^2\theta_i$, where $0\le\theta_i\lt\pi/2$.

8. For positive $a,b,c$, prove $a^3+b^3+c^3\ge a^2b+b^2c+c^2a$.

9. Let $f(x)=x^2+4x$. For $0\lt\varepsilon\lt1$, find a $\delta\gt0$ such that $|x+3|\lt\delta$ implies $|f(x)+3|\lt\varepsilon$.

10. For $x_n=(n^2-3)/(n^2-5n-1)$, prove $|x_n-1|\le14/n$ for $n\ge12$, then give an explicit $N$ such that $|x_n-1|\le\varepsilon_0$ for all $n\ge N$, where $\varepsilon_0=2^{-1025}$.

Solutions

Solution · Solution 1

The base case is $3/8\lt1/\sqrt7$, because both sides are positive and $63\lt64$. Suppose the claim holds for $k\ge2$. Multiplication by the next positive factor reduces the induction step to

$$\frac{2k+1}{(2k+2)\sqrt{3k+1}}\lt\frac1{\sqrt{3k+4}}.$$

After squaring positive quantities, the right side exceeds the left because $(2k+2)^2(3k+1)-(2k+1)^2(3k+4)=k>0$. This proves the step and hence the result.

Solution · Solution 2

Factor the first difference:

$$a^{n+1}-a^nb-(ab^n-b^{n+1})=(a-b)(a^n-b^n)\gt0.$$

Because $a-b$ and $a^n-b^n$ have the same nonzero sign, this is positive. For the second claim, the base case is equivalent to $(a-b)^2\gt0$. If it holds at $k$, then the positive difference

$$a b^k((k+1)a-kb)-b^{k+1}((k+2)a-(k+1)b) =(k+1)b^k(a-b)^2$$

places the desired left side below $a$ times the induction-hypothesis left side, hence below $a^{k+2}$. Finally set $a=n(n+2)$ and $b=(n+1)^2$, then divide the resulting positive factors to obtain $(1+1/n)^n\lt(1+1/(n+1))^{n+1}$.

Solution · Solution 3

For $n=1$ there is equality. If $(1+x)^k\ge1+kx$, then $1+x\gt0$, so

$$(1+x)^{k+1}\ge(1+kx)(1+x) =1+(k+1)x+kx^2\ge1+(k+1)x.$$

Thus induction proves the claim. Equality holds for every $n$ when $x=0$, and also for every admissible $x$ when $n=1$; for $n\gt1$, equality requires $x=0$.

Solution · Solution 4

On $x\gt1$, multiplying by $x-1\gt0$ gives $x\ge3$. On $x\lt1$, multiplication reverses the sign and gives $x\le3$, so the entire interval $x\lt1$ remains. Alternatively, after imposing $x\ne1$, multiplication by $(x-1)^2\gt0$ gives $(x-1)(x-3)\ge0$. Both methods yield $(-\infty,1)\cup[3,\infty)$.

Solution · Solution 5

Split the first problem at $2$. On $x\lt2$, it becomes $(x+2)(x-1)\gt0$; on $x\ge2$, the resulting quadratic is always positive. The answer is $x\lt-2$ or $x\gt1$. For the second problem, $x+1$ must be negative, so $x\lt-1$; splitting again at $-2$ leaves only $-3/2\lt x\lt-1$.

Solution · Solution 6

The three positive terms $a/b,b/c,c/a$ have product $1$, so their arithmetic mean is at least $1$. Equality requires $a/b=b/c=c/a=1$, hence $a=b=c$. Applying AM-GM to $1/a_1,\ldots,1/a_n$ gives $1/H_n\ge1/G_n$; positivity permits reciprocation, yielding $H_n\le G_n$.

Solution · Solution 7

Apply Cauchy-Schwarz to $a_i/\sqrt{b_i}$ and $\sqrt{b_i}$ to obtain Engel's bound. With the stated trigonometric substitution, every cosine is positive, so it gives

$$\sum_{i=1}^n\frac1{\cos^2\theta_i} \ge n+ \frac{(\sum_{i=1}^n\sin\theta_i)^2}{\sum_{i=1}^n\cos^2\theta_i}.$$

Here the identity $1/\cos^2\theta_i=1+\tan^2\theta_i$ supplies the displayed form.

Solution · Solution 8

By two-variable AM-GM, $2a^3+b^3\ge3a^2b$, $2b^3+c^3\ge3b^2c$, and $2c^3+a^3\ge3c^2a$. Adding and dividing by $3$ gives the result. Equality in all three comparisons requires $a=b=c$.

Solution · Solution 9

If $|x+3|\lt1$, then $|x+1|\le|x+3|+2\lt3$, and therefore $|f(x)+3|=|x+3||x+1|\le3|x+3|$. Take $\delta=\varepsilon/3$; since $0\lt\varepsilon\lt1$, the preliminary condition is satisfied, and the final estimate is strictly less than $\varepsilon$.

Solution · Solution 10

For $n\ge12$, $|5n-2|\le7n$ and $|n^2-5n-1|\ge n^2-6n\ge n^2/2$. Hence $|x_n-1|\le14/n$. Taking $N=14\cdot2^{1025}$ gives $14/N=2^{-1025}$ and therefore the required estimate for all $n\ge N$.