Cauchy-Schwarz and triangle inequalities

Control the cross term in a squared length

The inner product gives formulas for length and angle. The next question is: what restrictions do those formulas obey?

Two answers dominate the subject:

• the Cauchy-Schwarz inequality controls the size of an inner product by the lengths of the two vectors;
• the triangle inequality says the direct path from $0$ to $u+v$ is never longer than taking the two steps through $u$ and then $u+v$.

These inequalities explain why Euclidean geometry can be expressed algebraically.

The difficulty appears whenever a norm-square is expanded. For example, $\|u+v\|^2$ contains the signed cross term $2\langle u,v\rangle$, whereas a length must be nonnegative and should be controlled without knowing the coordinates of either vector. Cauchy-Schwarz supplies exactly that control. It then turns the algebraic expansion into the triangle inequality, and the equality conditions record whether the two vectors lie on the same ray or on opposite rays. The same estimates also show that small changes in a vector produce small changes in its length and in any fixed inner product.

There is also a structural reason to prove the results in this order. From the inner-product axioms, nonnegativity and homogeneity of the induced length are immediate, but the triangle inequality is not. Only after Cauchy-Schwarz controls the cross term do all norm axioms follow, and only then does $d(x,y)=\|x-y\|$ become a genuine metric. Thus this section is not merely a collection of useful estimates: it completes the logical passage from an inner product to length, distance, and angle.

The real inner-product setting

Definition

Real inner-product setting

Throughout this note, $V$ is a real inner-product space. Thus $\langle\cdot,\cdot\rangle$ is real-valued, bilinear, symmetric, and positive definite. The induced norm is

$$\|x\|=\sqrt{\langle x,x\rangle},$$

so $\|x\|\ge0$, with equality exactly when $x=0$. The standard model is $V=\mathbb{R}^m$ with the dot product, but the proofs below use only the real inner-product axioms and therefore apply to every real inner-product space.

Two vectors $v,w$ are linearly dependent when there are real numbers $a,b$, not both zero, such that $av+bw=0$. Under this definition, every pair containing the zero vector is linearly dependent. This convention is essential when the equality cases are stated without separately excluding zero vectors. Complex inner-product spaces require conjugate-linearity and a different discussion of angles; they are not being silently included here.

Cauchy-Schwarz inequality

Theorem

Cauchy-Schwarz inequality

For all vectors $v,w\in V$,

$$|\langle v,w\rangle|\le\|v\|\,\|w\|.$$

Equality holds if and only if $v$ and $w$ are linearly dependent.

The inequality says an inner product can never be larger in magnitude than the product of the two lengths. For nonzero vectors $v$ and $w$, the quotient below lies in $[-1,1]$, so their angle is well-defined by

$$\theta= \arccos\!\left( \frac{\langle v,w\rangle}{\|v\|\,\|w\|} \right)\in[0,\pi].$$

Equivalently, $\cos\theta=\langle v,w\rangle/(\|v\|\,\|w\|)$. If either vector is zero, the denominator vanishes, so this formula does not define an angle. For nonzero vectors, $\theta=0$ means they lie on the same ray, $\theta=\pi/2$ means they are orthogonal, and $\theta=\pi$ means they lie on opposite rays.

This definition depends only on direction. Multiplying either vector by a positive scalar leaves the quotient unchanged; multiplying exactly one vector by a negative scalar reverses its direction and changes the angle from $\theta$ to $\pi-\theta$. At the two endpoints, the quotient is $1$ for positive proportionality and $-1$ for negative proportionality. The absolute value in Cauchy-Schwarz deliberately treats both endpoints as equality cases, whereas the triangle inequality will distinguish them. A zero vector has no direction to normalize, which is the conceptual reason—besides division by zero—that no angle is assigned to it here.

Proof

The quadratic-polynomial proof

First suppose $w=0$. Both sides of Cauchy-Schwarz are then $0$, so equality is immediate; moreover, a pair containing the zero vector is linearly dependent. Thus both the inequality and the stated equality condition hold in this case.

Now suppose $w\ne0$ and consider the function

$$F(t)=\|tw-v\|^2.$$

Since a norm-square is always nonnegative, $F(t)\ge0$ for every real $t$. Expanding gives

$$F(t)=t^2\langle w,w\rangle-2t\langle v,w\rangle+\langle v,v\rangle.$$

Because its leading coefficient is $\|w\|^2>0$, this is a genuine quadratic polynomial in $t$. It is nonnegative for every real $t$, so its discriminant is nonpositive:

$$4\langle v,w\rangle^2-4\|v\|^2\|w\|^2\le0.$$

Therefore

$$|\langle v,w\rangle|\le\|v\|\,\|w\|.$$

In the case $w\ne0$, equality in Cauchy-Schwarz occurs exactly when the discriminant is zero, equivalently when $F(t)$ has a real root. Such a root means $tw-v=0$ for some real $t$, so $v$ is a scalar multiple of $w$. Conversely, if $v$ and $w$ are linearly dependent, then $v=tw$ for some real $t$, and direct substitution gives equality. Together with the case $w=0$, this proves the equality condition, including all cases in which one or both vectors are zero.

Projection meaning of the quadratic

The same calculation explains why the proof is geometrically natural. When $w\ne0$, choose

$$t_0=\frac{\langle v,w\rangle}{\|w\|^2}, \qquad r=v-t_0w.$$

Then

$$\langle r,w\rangle =\langle v,w\rangle-t_0\|w\|^2=0,$$

so $t_0w$ is the component of $v$ along the line spanned by $w$, while $r$ is the orthogonal residual. Expanding the orthogonal decomposition $v=t_0w+r$ gives

$$\|v\|^2 =\|t_0w\|^2+\|r\|^2 =\frac{\langle v,w\rangle^2}{\|w\|^2}+\|r\|^2.$$

Since $\|r\|^2\ge0$, the component of $v$ along a line cannot be longer than $v$ itself; after rearranging, this is Cauchy-Schwarz. Equality occurs exactly when $r=0$, which says $v$ already lies on that line. Thus the discriminant argument, orthogonal projection, and the equality condition are three views of the same fact.

One can also see the exact minimum by completing the square:

$$F(t) =\|w\|^2\left(t-\frac{\langle v,w\rangle}{\|w\|^2}\right)^2 +\|v\|^2-\frac{\langle v,w\rangle^2}{\|w\|^2}.$$

The first term vanishes precisely at $t=t_0$. Therefore the final constant is the minimum value of $F$, and it must be nonnegative. This produces the same squared inequality as the discriminant argument. More importantly, it shows why a zero discriminant is not merely an algebraic coincidence: it says the minimum distance from $v$ to the line spanned by $w$ is zero. In that case the orthogonal residual disappears and the vectors are linearly dependent.

A nonnegative quadratic gives Cauchy–Schwarz, whose bound on the cross term then gives the triangle inequality.

Worked example

Check Cauchy-Schwarz numerically

Let

$$v= \begin{bmatrix} 1\\2\\3 \end{bmatrix}, \qquad w= \begin{bmatrix} 4\\5\\6 \end{bmatrix}.$$

Then

$$\langle v,w\rangle=32, \qquad \|v\|=\sqrt{14}, \qquad \|w\|=\sqrt{77}.$$

So

$$|\langle v,w\rangle|=32, \qquad \|v\|\,\|w\|=\sqrt{14\cdot77}=\sqrt{1078}\approx32.83.$$

Indeed,

$$32\le32.83.$$

The normalized inner product is $32/\sqrt{1078}\approx0.9746$, so the angle is approximately $0.226$ radians. Because the normalized value is strictly less than $1$, the vectors are close in direction but are not linearly dependent.

Triangle inequality

Theorem

Triangle inequality

For all vectors $u,v\in V$,

$$\|u+v\|\le\|u\|+\|v\|.$$

Equality holds if and only if at least one of $u$ and $v$ is a nonnegative scalar multiple of the other, including when one vector is zero.

The geometric reading is direct: in Euclidean space, one side of a triangle is never longer than the sum of the other two sides.

Proof

The equality case must survive both inequalities

There are two successive equality gaps to track:

$$\langle u,v\rangle\le|\langle u,v\rangle|\le\|u\|\,\|v\|.$$

Equality in the first gap controls the sign of the inner product; equality in the second controls linear dependence. The zero-vector cases are kept separate, so no normalization is needed there.

Start with

$$\|u+v\|^2 =\langle u+v,u+v\rangle =\|u\|^2+2\langle u,v\rangle+\|v\|^2.$$

By Cauchy-Schwarz,

$$\langle u,v\rangle\le|\langle u,v\rangle|\le\|u\|\,\|v\|.$$

Therefore

$$\|u+v\|^2\le\|u\|^2+2\|u\|\,\|v\|+\|v\|^2=(\|u\|+\|v\|)^2.$$

Taking square roots gives

$$\|u+v\|\le\|u\|+\|v\|.$$

For equality, both estimates in the chain must be tight. The first requires $\langle u,v\rangle=|\langle u,v\rangle|$, so the inner product is nonnegative. The Cauchy-Schwarz bound is tight exactly when $u$ and $v$ are linearly dependent. If both vectors are nonzero, write $v=\lambda u$; then $\langle u,v\rangle=\lambda\|u\|^2\ge0$ forces $\lambda\ge0$. If one vector is zero, equality is immediate, and the zero vector is a nonnegative scalar multiple of the other. Conversely, vectors satisfying this nonnegative-multiple condition give equality by direct substitution. Thus nonzero vectors pointing in opposite directions give equality in Cauchy-Schwarz, but not in the triangle inequality.

For nonzero $u$ and $v$, the angle definition makes the geometric connection explicit:

$$\|u+v\|^2 =\|u\|^2+\|v\|^2+2\|u\|\,\|v\|\cos\theta.$$

Cauchy-Schwarz is the assertion that $-1\le\cos\theta\le1$; using the upper endpoint gives the triangle bound. Equality at that endpoint requires $\theta=0$, not $\theta=\pi$. This is why same-line vectors with opposite directions satisfy equality in Cauchy-Schwarz but fail to satisfy equality in the triangle inequality. The algebraic sign condition and the geometric ray condition are therefore equivalent descriptions.

There is a useful quantitative version of this picture. Normalize nonzero vectors by setting $\widehat u=u/\|u\|$ and $\widehat v=v/\|v\|$. Then they are unit vectors and

$$\|\widehat u-\widehat v\|^2 =2-2\langle\widehat u,\widehat v\rangle =2-2\cos\theta =4\sin^2\!\left(\frac{\theta}{2}\right).$$

Because $\theta/2\in[0,\pi/2]$, taking the nonnegative square root gives $\|\widehat u-\widehat v\|=2\sin(\theta/2)$. Thus a small angle means the two normalized vectors are close in norm, while angle $\pi$ puts them at the maximum possible distance $2$. Similarly, $\|\widehat u+\widehat v\|=2\cos(\theta/2)$, so the upper triangle bound $2$ is attained only at angle zero. These identities translate the equality conditions into exact distances on the unit sphere.

Theorem

Reverse triangle inequality

For all vectors $u,v\in V$,

$$\|u-v\|\ge\bigl|\|u\|-\|v\|\bigr|.$$

Equality holds if and only if at least one of $u$ and $v$ is a nonnegative scalar multiple of the other, with zero-vector cases included.

Apply the triangle inequality to $u=(u-v)+v$ and to $v=(v-u)+u$. This gives both $\|u\|-\|v\|\le\|u-v\|$ and $\|v\|-\|u\|\le\|u-v\|$, which prove the displayed bound. If, for example, $\|u\|\ge\|v\|$ and equality holds, then the triangle estimate for $u=(u-v)+v$ must itself be an equality. Its equality condition puts $u-v$ and $v$ on the same nonnegative ray, and hence puts $u$ and $v$ on that ray as well. The other norm ordering is symmetric. Conversely, the bound is an equality whenever $u$ and $v$ satisfy the stated nonnegative-multiple condition, including when either is zero.

This version is often easier to use when you want a lower bound instead of an upper bound.

It can also be written in the symmetric form

$$\bigl|\|u\|-\|v\|\bigr|\le\|u-v\|.$$

Thus the norm function is 1-Lipschitz for the distance $d(u,v)=\|u-v\|$: changing a vector by at most $\varepsilon$ changes its length by at most $\varepsilon$. In particular, if $x_n$ approaches $x$ in norm, then $|\|x_n\|-\|x\||\le\|x_n-x\|$ shows directly that $\|x_n\|$ approaches $\|x\|$. This analytic consequence is the same reverse triangle inequality, not an additional assumption about coordinates.

Repeated application also gives the finite-sum form

$$\left\|\sum_{k=1}^{n}x_k\right\| \le\sum_{k=1}^{n}\|x_k\|.$$

This follows by induction: separate the last summand, apply the two-vector triangle inequality, and then apply the induction hypothesis to the preceding sum. Combining it with norm homogeneity yields

$$\left\|\sum_{k=1}^{n}a_kx_k\right\| \le\sum_{k=1}^{n}|a_k|\,\|x_k\|,$$

which is the standard way to control the accumulated size of several vector contributions without expanding coordinates.

The equality case has a useful many-vector form. For a positive integer $p$ and vectors $x_1,\ldots,x_p$, expanding both squares gives

$$\left(\sum_{i=1}^{p}\|x_i\|\right)^2- \left\|\sum_{i=1}^{p}x_i\right\|^2 =2\sum_{1\le i<j\le p} \left(\|x_i\|\,\|x_j\|-\langle x_i,x_j\rangle\right)\ge0$$

Every pairwise gap is nonnegative by Cauchy-Schwarz together with $\langle x_i,x_j\rangle\le|\langle x_i,x_j\rangle|$. Thus equality holds exactly when every pairwise gap is zero. For $p=1$ this is immediate. If all vectors are zero, it is also immediate. Otherwise choose a nonzero $w=x_k$; equality with $w$ and the two-vector equality case force every nonzero $x_i$ to be a positive scalar multiple of $w$, while zero vectors are allowed. Conversely, vectors on one positive ray make the norm sum additive. Thus equality means one common positive ray, not merely one line.

The same result is the polygonal-path inequality. For points $t_0,\ldots,t_p$, let $x_i=t_i-t_{i-1}$. Since

$$\sum_{i=1}^{p}x_i=t_p-t_0$$

the finite-sum inequality gives

$$\|t_p-t_0\|\le\sum_{i=1}^{p}\|t_i-t_{i-1}\|$$

Equality holds exactly when all nonzero steps lie on the same positive ray; zero steps are allowed, and an all-zero path is valid. In geometric terms, the path moves in one direction without backtracking. If $t_p=t_0$, equality therefore forces every step to be zero.

Cauchy-Schwarz provides the corresponding stability estimate for inner products. For a fixed vector $z$,

$$|\langle x,z\rangle-\langle y,z\rangle| =|\langle x-y,z\rangle| \le\|x-y\|\,\|z\|.$$

If both arguments change, add and subtract $\langle x',y\rangle$ and apply Cauchy-Schwarz twice to obtain

$$|\langle x,y\rangle-\langle x',y'\rangle| \le\|x-x'\|\,\|y\|+\|x'\|\,\|y-y'\|.$$

These bounds explain, in a coordinate-free way, why norms and inner products are stable under small perturbations.

Equality cases matter

The equalities are not side remarks. They tell you exactly when the estimate is tight.

• For nonzero vectors, equality in Cauchy-Schwarz means the vectors point in the same or opposite direction. More generally, equality means they are linearly dependent, which also includes every pair containing a zero vector.
• For nonzero vectors, equality in the triangle inequality means they point in the same direction with a nonnegative scalar factor. A pair containing a zero vector is also an equality case.

Worked example

When triangle inequality becomes equality

Let

$$u= \begin{bmatrix} 2\\0 \end{bmatrix}, \qquad v= \begin{bmatrix} 3\\0 \end{bmatrix}.$$

Then $v=\frac32u$, so $v$ is a nonnegative scalar multiple of $u$. Hence

$$\|u+v\|=\left\| \begin{bmatrix} 5\\0 \end{bmatrix} \right\|=5 =2+3 =\|u\|+\|v\|.$$

So equality holds.

Worked example

Cauchy-Schwarz equality does not determine the triangle sign

Let

$$u=(2,0), \qquad v=(-3,0).$$

The vectors are linearly dependent because $v=-\frac32u$. Consequently, Cauchy-Schwarz is an equality, as the direct calculation confirms:

$$|\langle u,v\rangle|=|-6|=6 =\|u\|\,\|v\|=2\cdot3.$$

For the triangle inequality, however,

$$\|u+v\|=\|(-1,0)\|=1 <5=\|u\|+\|v\|.$$

The inequality is strict because the scalar $-3/2$ is negative. This example protects the sign condition: linear dependence alone is enough for equality in Cauchy-Schwarz, but triangle equality requires a nonnegative scalar multiple.

Common mistake

Common mistake

Equality in Cauchy-Schwarz is not the same as orthogonality

Orthogonality means $\langle v,w\rangle=0$. Equality in Cauchy-Schwarz means

$$|\langle v,w\rangle|=\|v\|\,\|w\|.$$

For nonzero $v$ and $w$, equality means the vectors are linearly dependent and cannot also be orthogonal. If either vector is zero, however, the vectors are both orthogonal and linearly dependent, so equality does hold in that degenerate case.

Common mistake

Do not lose the sign or the nonzero hypothesis

The absolute-value equality $|\langle u,v\rangle|=\|u\|\,\|v\|$ allows both positive and negative scalar multiples. In the triangle proof, one also needs $\langle u,v\rangle=|\langle u,v\rangle|$; omitting this step incorrectly admits opposite directions. Separately, the angle formula divides by both norms, so it may be used only when both vectors are nonzero. The inequalities themselves remain valid when a vector is zero, but the angle does not become defined by assigning a convenient value.

Quick check

Checkpoint

What does Cauchy-Schwarz say about $|\langle v,w\rangle|$?

State the inequality directly.

Solution · Answer

$$|\langle v,w\rangle|\le\|v\|\,\|w\|.$$

Checkpoint

If $u$ and $v$ point in exactly the same direction, can equality hold in the triangle inequality?

Assume one is a nonnegative scalar multiple of the other.

Solution · Answer

Yes. That is exactly the equality case for the triangle inequality.

Checkpoint

If $v\perp w$, what does Cauchy-Schwarz reduce to?

Use $\langle v,w\rangle=0$.

Solution · Answer

It becomes $0\le\|v\|\,\|w\|$, which is automatically true.

Exercises

Checkpoint

Use Cauchy-Schwarz to show that $|\langle (1,2),(3,4)\rangle| \le \|(1,2)\|\,\|(3,4)\|$.

Compute both sides numerically.

Solution · Guided solution

The left-hand side is

$$|1\cdot3+2\cdot4|=11.$$

The right-hand side is

$$\sqrt{1^2+2^2}\sqrt{3^2+4^2} =\sqrt5\cdot5=5\sqrt5\approx11.18.$$

So indeed $11\le5\sqrt5$.

Checkpoint

Find $\|u+v\|$ and compare it with $\|u\|+\|v\|$ for $u=(1,0)$ and $v=(0,1)$.

This is the right-angle case.

Solution · Guided solution

$$u+v=(1,1), \qquad \|u+v\|=\sqrt2.$$

Also,

$$\|u\|+\|v\|=1+1=2.$$

So $\sqrt2\le2$, and the inequality is strict because the vectors are not nonnegative scalar multiples of each other.

Checkpoint

Use the reverse triangle inequality to bound $\|u-v\|$ from below when $\|u\|=7$ and $\|v\|=3$.

Apply the formula directly.

Solution · Guided solution

$$\|u-v\|\ge|\|u\|-\|v\||=|7-3|=4.$$

Keep the bound and its equality case together

Cauchy-Schwarz is the fundamental inner-product estimate: $|\langle v,w\rangle|\le\|v\|\,\|w\|$. In a real inner-product space it both bounds the component of one vector along another and, for two nonzero vectors, guarantees that the angle $\theta=\arccos(\langle v,w\rangle/(\|v\|\,\|w\|))$ lies in $[0,\pi]$. Its equality condition is linear dependence, including all zero-vector cases and both signs of proportionality.

Controlling the cross term in $\|u+v\|^2$ yields the triangle inequality. Equality there is more restrictive: one vector must be a nonnegative scalar multiple of the other. Applying the same inequality to a difference yields the reverse triangle inequality and the stability estimate $|\|u\|-\|v\||\le\|u-v\|$. The proof chain is therefore inner-product positivity, then Cauchy-Schwarz, then triangle and reverse triangle inequalities; the equality cases distinguish a line from a ray.

Related notes

Read 9.1 Inner products, norms, and angles first, because these inequalities are built directly from the inner-product and norm formulas.

The coefficient formulas in 9.2 Orthogonal sets and orthonormal bases and the geometry of 9.3 Gram-Schmidt orthogonalization are established from inner-product algebra and orthogonality. The inequalities here provide bounds for those coefficients and residual lengths.