Elementary inequalities: identities, domains and equality cases

Learning goals and strategy

Prove inequalities by transforming the difference, use nonnegative squares and exact identities, track denominator signs, and determine equality conditions. The objective is not to memorize isolated bounds. A good proof identifies why a quantity is nonnegative and which assumptions make the transformation reversible.

Three useful approaches are: write the difference between the two sides as a sum of squares; recognize a Cauchy–Schwarz or arithmetic–geometric mean structure; or introduce a parameter and force its quadratic expression to stay nonnegative. For every chain of inequalities, equality in the final result requires equality at every step that could be strict.

1. The first inequality is an identity plus a sign

For nonnegative reals $a,b$,

$$a+b-2\sqrt{ab}=(\sqrt a-\sqrt b)^2\ge0.$$

Thus $(a+b)/2\ge\sqrt{ab}$, with equality exactly when $a=b$. The radicands are defined because the inputs are nonnegative. If instead $a,b>0$, dividing by their positive product gives equivalent reciprocal inequalities; a denominator of unknown sign must not be treated the same way.

For nonzero real $a,b$ there is an exact identity

$$(a+b)\left(\frac1a+\frac1b\right)-4=\frac{(a-b)^2}{ab}.$$

If $ab>0$, the expression on the left is nonnegative; if $ab<0$, it is nonpositive. The same algebra produces opposite inequality directions because of the denominator. Positive inputs are a convenient sufficient hypothesis, but the exact condition for this nonnegative remainder is the sign of $ab$, not a vague appeal to “squaring makes everything positive.”

2. A full Cauchy–Schwarz proof and equality condition

For a positive integer $n$ and real vectors $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$, the identity

$$\left(\sum_i u_i^2\right)\left(\sum_i v_i^2\right)-\left(\sum_i u_iv_i\right)^2 =\sum_{i<j}(u_iv_j-u_jv_i)^2$$

follows by expansion: diagonal terms cancel and each off-diagonal pair contributes $u_i^2v_j^2+u_j^2v_i^2-2u_iu_jv_iv_j$. Every remaining term is a square, so

$$\left|\sum_i u_iv_i\right|\le\sqrt{\sum_i u_i^2}\sqrt{\sum_i v_i^2}.$$

Equality holds exactly when every cross term $u_iv_j-u_jv_i$ vanishes. If $v=0$, equality holds for every $u$. If some $v_j\ne0$, the identities give $u_i=(u_j/v_j)v_i$ for every $i$, so $u$ is a scalar multiple of $v$. Conversely scalar multiples make every cross term zero. Thus the equality condition, including the zero-vector case, has been proved rather than guessed from a picture.

Method interpretation. The identity method exposes every lost nonnegative term. A parameter method would instead start with $\sum_i(u_i-tv_i)^2\ge0$ for all real $t$ and minimize its quadratic. Both are valid, but the identity often makes equality and stability particularly transparent.

3. Weighted squares reveal the parameter range

For real $a,b$ and a real parameter $t$,

$$(1-t)a^2+tb^2-[(1-t)a+tb]^2=t(1-t)(a-b)^2.$$

For $0\le t\le1$, this is nonnegative, proving the convex-square inequality. Equality occurs when $t=0$, $t=1$, or $a=b$. For $t<0$ or $t>1$ and $a\ne b$, the right side is negative and the inequality reverses strictly. Calling $t$ a “weight” without imposing its allowed range would conceal an essential hypothesis.

As another parameter example, $x^2+2tx+1=(x+t)^2+1-t^2$ is nonnegative for every real $x$ exactly when $|t|\le1$. Sufficiency follows from both terms being nonnegative; necessity follows by evaluating at the deliberately chosen point $x=-t$. An existential counterexample can depend on the parameter it is designed to defeat.

4. Complex triangle and reverse-triangle inequalities

First recall the modulus product identity:

Conjugation preserves sums and products by expansion. Hence

$$|zw|^2=(zw)\overline{(zw)}=(z\overline z)(w\overline w)=|z|^2|w|^2.$$

Both sides' square roots are nonnegative, giving $|zw|=|z||w|$. The modulus is a nonnegative real measure of size, not an order relation between complex numbers.

For $\xi=x+iy$ with $x,y\in\mathbb R$, $|\xi|=\sqrt{x^2+y^2}\ge|x|\ge x=\operatorname{Re}\xi$.

For complex $z,w$, expansion gives

$$|z+w|^2=|z|^2+|w|^2+2\operatorname{Re}(z\overline w) \le |z|^2+|w|^2+2|z||w|=(|z|+|w|)^2.$$

The inequality uses $\operatorname{Re}\xi\le|\xi|$ and the already proved modulus product identity. Both sides to be square-rooted are nonnegative, so $|z+w|\le|z|+|w|$.

If neither input is zero, equality holds exactly when $z\overline w$ is nonnegative real, equivalently $z=\lambda w$ for a real $\lambda>0$. Allowing a zero input extends this to vectors on the same nonnegative ray. The equality condition is not merely “both numbers are real”: opposite-sign reals usually give strict triangle inequality.

Applying the triangle inequality to $z=(z-w)+w$ gives $|z|-|w|\le|z-w|$. Interchanging $z,w$ gives the other direction, hence

$$\bigl||z|-|w|\bigr|\le|z-w|.$$

The absolute value outside the modulus difference is necessary because either modulus can be larger. This reverse-triangle bound often proves continuity of modulus directly: nearby inputs have nearby sizes.

5. Equality conditions can produce stability estimates

Suppose $a,b\ge0$, $a+b=2$, and $ab\ge1-\varepsilon$ for $\varepsilon\ge0$. The exact identity

$$(a-b)^2=(a+b)^2-4ab=4-4ab\le4\varepsilon$$

gives $|a-b|\le2\sqrt\varepsilon$. Since $a-1=(a-b)/2$ and $b-1=(b-a)/2$, each is within $\sqrt\varepsilon$ of the equality value one. This does more than identify where a maximum occurs: a nearly maximal product forces the inputs quantitatively close to one another.

Likewise the complex identity

$$2(|z|^2+|w|^2)-|z+w|^2=|z-w|^2$$

gives $|z+w|\le\sqrt2\sqrt{|z|^2+|w|^2}$, with equality exactly when $z=w$. The identity is often cleaner than chaining triangle inequality with a real Cauchy bound and then separately reconciling their equality conditions. When a desired bound looks quadratic, first check whether an exact squared-distance identity is available.

Worked method choice

To prove $\sum_{j=1}^n1/a_j\ge n^2/\sum_{j=1}^na_j$ for positive $a_j$, a common denominator is unwieldy. Cauchy–Schwarz with $u_j=\sqrt{a_j}$ and $v_j=1/\sqrt{a_j}$ makes their dot product exactly $n$, so it immediately gives the desired bound after division by a positive sum. Equality requires $u_j=\lambda v_j$ for all $j$, hence all $a_j$ equal. The substitutions were chosen to simplify the dot product, not by pattern matching after the proof was finished.

To refute an unrestricted reciprocal inequality, on the other hand, a sign choice is more efficient than a general theorem. A single opposite-sign pair makes the exact remainder negative. A proof method should fit the goal: theorem construction, domain diagnosis and counterexample search are different tasks.

Graded proof and diagnostic tasks

1. Method choice. Prove $(\sum a_j)(\sum1/a_j)\ge n^2$ for positive reals and characterize equality, explaining why each square root and division is permitted.
2. Parameter range. Determine all real $t$ for which $(1-t)a^2+tb^2\ge[(1-t)a+tb]^2$ holds for every pair of real $a,b$.
3. Universal quantifier. Determine all real $t$ for which $x^2+2tx+1\ge0$ for every real $x$, and describe when equality can occur.
4. Sign audit. Diagnose a proof that multiplies a rational inequality by $ab$ without checking its sign. Use the exact reciprocal identity to state the valid conclusions for $ab>0$ and $ab<0$.
5. Complex equality. Prove $|z+w|^2\le2(|z|^2+|w|^2)$ by a squared-distance identity, and give its full equality condition including zero inputs.
6. Quantitative refinement. For $a,b\ge0$, $a+b=S>0$ and $ab\ge S^2/4-\delta$ with $\delta\ge0$, bound each distance from $S/2$. Explain why this is a stability statement rather than only an equality-case statement.

Full solutions

1. Put $u_j=\sqrt{a_j}$ and $v_j=1/\sqrt{a_j}$. Positivity makes all expressions defined. Their dot product is $n$, so Cauchy–Schwarz gives $n^2\le(\sum a_j)(\sum1/a_j)$. Equality occurs precisely when $\sqrt{a_j}=\lambda/\sqrt{a_j}$ for every $j$, so every $a_j=\lambda$; conversely equal positive inputs attain equality. No cancellation by zero or complex square-root branch is involved.
2. The difference is $t(1-t)(a-b)^2$. It is nonnegative for all pairs if $0\le t\le1$. If $t$ is outside that interval, choose $a=0,b=1$ to make the difference strictly negative. Thus exactly the closed unit interval works.
3. Completing the square gives $(x+t)^2+1-t^2$. It is nonnegative for all $x$ if $|t|\le1$. If $|t|>1$, choose $x=-t$ to get a negative value, proving necessity. Equality occurs only when $|t|=1$ and $x=-t$; when $|t|<1$, the expression is strictly positive everywhere.
4. Multiplying by a positive denominator preserves an inequality, a negative one reverses it, and a zero one makes the original reciprocal undefined. Since the exact difference from four is $(a-b)^2/(ab)$, the product is at least four when $ab>0$, with equality at $a=b$. For $ab<0$, it is strictly below four, because opposite-sign nonzero inputs cannot be equal and the numerator is positive. The unverified sign is the first invalid inference.
5. Expand $|z-w|^2=|z|^2+|w|^2-2\operatorname{Re}(z\overline w)$ and subtract the expansion for $|z+w|^2$ to obtain $2(|z|^2+|w|^2)-|z+w|^2=|z-w|^2\ge0$. Equality holds exactly when $z-w=0$, namely $z=w$, including the pair of zeros.
6. Since $(a-b)^2=S^2-4ab\le4\delta$, $|a-b|\le2\sqrt\delta$. The identities $a-S/2=(a-b)/2$ and $b-S/2=(b-a)/2$ give both distances at most $\sqrt\delta$. At $\delta=0$ this recovers exact equality. For positive $\delta$ it controls how far a near-equality configuration can move, which is the additional content of stability.

Summary and terms

An inequality identity expresses the difference of two sides exactly before assigning its sign. An equality case records when every nonnegative remainder vanishes. A stability bound quantifies closeness when the remainder is merely small. A domain condition ensures that roots, reciprocals and sign-sensitive multiplications are valid. Proof quality includes those conditions and equality descriptions, not only the final inequality symbol.