Cauchy–Schwarz and the Triangle Inequality

Derivation

Cauchy–Schwarz and the Triangle Inequality

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Statement

In any real or complex inner-product space \((V,\langle\cdot,\cdot\rangle)\) with induced norm \(\|x\|=\sqrt{\langle x,x\rangle}\), every pair of vectors \(u,v\in V\) satisfies the Cauchy–Schwarz inequality \(\left|\langle u,v\rangle\right|\le \|u\|\,\|v\|\), with equality if and only if \(u\) and \(v\) are linearly dependent. From this the norm obeys the triangle inequality \(\|u+v\|\le\|u\|+\|v\|\), so \(\|\cdot\|\) is a genuine norm.

Why it matters

Cauchy–Schwarz is the single inequality that converts an inner product — an algebraic bilinear form — into a geometry. It guarantees that \(\langle u,v\rangle/(\|u\|\|v\|)\) lies in \([-1,1]\), so the angle defined by \(\cos\theta=\langle u,v\rangle/(\|u\|\|v\|)\) makes sense, and it is what forces \(\sqrt{\langle x,x\rangle}\) to satisfy the triangle inequality and hence qualify as a distance.

Every corner of physics that uses length, overlap, or orthogonality rests on it: the normalisation and overlap integrals of quantum states, the Heisenberg uncertainty relation, least-squares fitting and projection, the stability of Fourier series in \(L^2\), and the variational bounds of the Rayleigh–Ritz method. Symmetry enters through the Hermitian structure of \(\langle\cdot,\cdot\rangle\) — it is the conjugate symmetry and positivity of the form that force the bound.

Assumptions

Positive-definiteness: \(\langle x,x\rangle\ge 0\) with equality only for \(x=0\).If dropped (an indefinite metric, as in Minkowski space), \(\|x\|^2\) can be negative or null for \(x\neq0\); the square root is not a norm and the inequality reverses for timelike vectors. Conjugate symmetry: \(\langle u,v\rangle=\overline{\langle v,u\rangle}\).Without it \(\langle x,x\rangle\) need not be real, so \(\|x\|=\sqrt{\langle x,x\rangle}\) is not a real length and the discriminant argument below has no meaning. Sesquilinearity: \(\langle u,\alpha v+\beta w\rangle=\alpha\langle u,v\rangle+\beta\langle u,w\rangle\) (linear in the second slot, conjugate-linear in the first).If the form is not linear (only, say, subadditive) the expansion of \(\|u-\lambda v\|^2\) into a quadratic in \(\lambda\) breaks, and the minimisation proof collapses. Completeness is not required.Cauchy–Schwarz holds in any inner-product space, not only a Hilbert space. Dropping completeness costs nothing here — only limits of sequences, not the inequality itself, would be affected.

Derivation

\[ 0\le \langle u-\lambda v,\;u-\lambda v\rangle,\qquad \lambda\in\mathbb{C}. \]

Positive-definiteness applied to the vector \(u-\lambda v\); true for every scalar \(\lambda\). A

\[ 0\le \langle u,u\rangle-\bar\lambda\,\langle v,u\rangle-\lambda\,\langle u,v\rangle+|\lambda|^2\langle v,v\rangle. \]

Expand by sesquilinearity: \(\langle u,\lambda v\rangle=\lambda\langle u,v\rangle\) and \(\langle\lambda v,u\rangle=\bar\lambda\langle v,u\rangle\). A

\[ \text{If } v=0:\ \langle u,v\rangle=0=\|u\|\,\|v\|\ \checkmark.\quad\text{Henceforth assume } v\neq0. \]

Dispose of the degenerate case so we may divide by \(\|v\|^2=\langle v,v\rangle>0\). A

\[ \lambda=\frac{\langle v,u\rangle}{\langle v,v\rangle}. \]

Choose \(\lambda\) as the projection coefficient of \(u\) onto \(v\) — the value minimising the right side. This is the geometric heart: subtract the component of \(u\) along \(v\). B

\[ 0\le \langle u,u\rangle-\frac{|\langle v,u\rangle|^2}{\langle v,v\rangle}-\frac{|\langle u,v\rangle|^2}{\langle v,v\rangle}+\frac{|\langle v,u\rangle|^2}{\langle v,v\rangle^2}\,\langle v,v\rangle. \]

Substitute the chosen \(\lambda\). With \(\langle u,v\rangle=\overline{\langle v,u\rangle}\), each mixed term is \(|\langle u,v\rangle|^2/\langle v,v\rangle\). C

\[ 0\le \langle u,u\rangle-\frac{|\langle u,v\rangle|^2}{\langle v,v\rangle}. \]

The last term simplifies to \(+|\langle u,v\rangle|^2/\langle v,v\rangle\); combined with the two negatives, \(-2X+X=-X\) with \(X=|\langle u,v\rangle|^2/\langle v,v\rangle\). Only conjugate symmetry was used. C

\[ |\langle u,v\rangle|^2\le \langle u,u\rangle\,\langle v,v\rangle=\|u\|^2\,\|v\|^2. \]

Multiply through by \(\langle v,v\rangle>0\) and rearrange. A

\[ \boxed{\ \left|\langle u,v\rangle\right|\le \|u\|\,\|v\|\ } \]

Take the non-negative square root. Equality in step 1 needs \(u-\lambda v=0\), i.e. \(u\parallel v\) (linear dependence). A

\[ \|u+v\|^2=\langle u+v,\,u+v\rangle=\|u\|^2+2\,\mathrm{Re}\langle u,v\rangle+\|v\|^2. \]

Expand by sesquilinearity; the cross terms \(\langle u,v\rangle+\langle v,u\rangle=2\,\mathrm{Re}\langle u,v\rangle\). A

\[ \mathrm{Re}\langle u,v\rangle\le|\langle u,v\rangle|\le\|u\|\,\|v\|. \]

The real part never exceeds the modulus; then apply Cauchy–Schwarz (step 8). A

\[ \|u+v\|^2\le\|u\|^2+2\|u\|\,\|v\|+\|v\|^2=\big(\|u\|+\|v\|\big)^2. \]

Substitute the bound of step 10 into step 9; recognise the perfect square. B

\[ \boxed{\ \|u+v\|\le\|u\|+\|v\|\ } \]

Both sides are non-negative, so the square root preserves the inequality. The triangle inequality holds; \(\|\cdot\|\) is a norm. A

Result

\[ \left|\langle u,v\rangle\right|\le \|u\|\,\|v\|\qquad\Longrightarrow\qquad \|u+v\|\le\|u\|+\|v\| \]

Reading. The overlap of two vectors can never exceed the product of their lengths; the closer \(|\langle u,v\rangle|\) is to \(\|u\|\|v\|\), the more nearly parallel they are, reaching equality exactly when one is a scalar multiple of the other. The triangle inequality then says a straight path is never longer than a detour: the length of a sum is at most the sum of the lengths.

Units check. The inner product carries units of (unit of \(u\))\(\times\)(unit of \(v\)); so does \(\|u\|\,\|v\|=\sqrt{\langle u,u\rangle}\,\sqrt{\langle v,v\rangle}\). Both sides of Cauchy–Schwarz match. In the triangle inequality \(u\) and \(v\) share the same units and every term is a length in those units — dimensionally consistent.

Limiting cases

Parallel vectors \(u=\alpha v\): \(|\langle u,v\rangle|=|\alpha|\,\|v\|^2=\|u\|\|v\|\); equality holds, and \(\|u+v\|=\|u\|+\|v\|\) when \(\alpha>0\).
Orthogonal vectors \(\langle u,v\rangle=0\): the bound is slack (\(0\le\|u\|\|v\|\)); the triangle inequality tightens to Pythagoras, \(\|u+v\|^2=\|u\|^2+\|v\|^2\).
Antiparallel \(u=-\alpha v,\ \alpha>0\): Cauchy–Schwarz still saturates, but \(\|u+v\|=\big|\,\|u\|-\|v\|\,\big|\), the reverse-triangle extreme.
One vector zero \(v=0\): both sides vanish; trivially equal.
Real two-dimensional case: reduces to \((u_1v_1+u_2v_2)^2\le(u_1^2+u_2^2)(v_1^2+v_2^2)\), i.e. \(\cos^2\theta\le1\).

Breaks when

Indefinite metrics. In Minkowski space with signature \((+,-,-,-)\) the form is not positive-definite. For two timelike vectors the reverse Cauchy–Schwarz \(|\langle u,v\rangle|\ge\|u\|\|v\|\) holds, and the ordinary triangle inequality reverses — the reverse triangle inequality is exactly what underlies the twin paradox.
Non-inner-product norms. For an \(L^p\) norm with \(p\neq2\) there is no inner product generating it; the parallelogram law fails, so the discriminant proof does not apply. The triangle inequality still holds (Minkowski's inequality) but must be proved via Hölder, not Cauchy–Schwarz.
Semi-inner products. If \(\langle x,x\rangle=0\) is allowed for some \(x\neq0\) (a seminorm), the equality condition weakens and \(\|\cdot\|\) fails to separate points; one must quotient by the null space before it becomes a norm.

Failure modes

Dropping the modulus. Writing \(\langle u,v\rangle\le\|u\|\|v\|\) in the complex case: the left side is generally complex, so the ordering is meaningless. The bound is on \(|\langle u,v\rangle|\).
Forgetting conjugation. Expanding \(\|u-\lambda v\|^2\) as if \(\langle\lambda v,u\rangle=\lambda\langle v,u\rangle\) in a complex space. It is \(\bar\lambda\langle v,u\rangle\); missing the bar yields a false, stronger inequality.
Confusing equality with orthogonality. Equality in Cauchy–Schwarz means parallel; students wrongly attach it to the orthogonal case, where the bound is in fact loosest.
Squaring an unsigned inequality. Deriving the triangle inequality by squaring without checking both sides are non-negative — legitimate here, but a habit that fails elsewhere.
Treating it as the triangle inequality itself. Cauchy–Schwarz bounds an inner product, not \(\|u+v\|\); the triangle inequality is a corollary, not the same statement.

Discussion

The proof is a disguised minimisation. Choosing \(\lambda=\langle v,u\rangle/\langle v,v\rangle\) subtracts from \(u\) exactly its projection onto \(v\), leaving the orthogonal residual \(r=u-\lambda v\). Cauchy–Schwarz is then just the statement \(\|r\|^2\ge0\): the length that survives after removing the shared component cannot be negative. Equality is the case where nothing survives — \(u\) lay entirely along \(v\). The inequality and the geometry of projection are the same fact seen twice.

Conceptually, Cauchy–Schwarz is what makes an inner-product space a metric space. The triangle inequality is the axiom that turns \(\|u-v\|\) into a distance; without it, convergence, continuity, and compactness would have no consistent meaning. Everything analytic built on Hilbert space — completeness, the projection theorem, Riesz representation — presupposes this bound.

In quantum mechanics the inequality is structural, not decorative. Set \(u=\Delta\hat A\,|\psi\rangle\) and \(v=\Delta\hat B\,|\psi\rangle\) with \(\Delta\hat A=\hat A-\langle\hat A\rangle\). Cauchy–Schwarz \(|\langle u,v\rangle|^2\le\langle u,u\rangle\langle v,v\rangle\) reads \(\langle(\Delta\hat A)^2\rangle\langle(\Delta\hat B)^2\rangle\ge|\langle\Delta\hat A\,\Delta\hat B\rangle|^2\). Taking the imaginary part of the cross term supplies the commutator, giving the Robertson–Schrödinger relation \(\sigma_A\,\sigma_B\ge\tfrac12\big|\langle[\hat A,\hat B]\rangle\big|\). The Heisenberg bound is a direct child of the projection residual being non-negative — an expression of the positivity and symmetry of the Hermitian inner product.

Common misconceptions. Cauchy–Schwarz is not a claim that angles are less than \(90^\circ\); it is what defines the angle by guaranteeing the cosine lies in \([-1,1]\). And it is not equivalent to the triangle inequality — the triangle inequality follows from it, yet many normed spaces satisfy the triangle inequality while possessing no inner product at all.

Worked examples

\[ \textbf{Overlap of two qubit states.}\quad |\psi\rangle=\tfrac{1}{\sqrt2}\big(|0\rangle+|1\rangle\big),\ \ |\phi\rangle=\cos\alpha\,|0\rangle+\sin\alpha\,|1\rangle. \]

Both are normalised: \(\||\psi\rangle\|=\||\phi\rangle\|=1\). Cauchy–Schwarz forces \(|\langle\psi|\phi\rangle|\le1\). A

\[ \langle\psi|\phi\rangle=\tfrac{1}{\sqrt2}\big(\cos\alpha+\sin\alpha\big)=\tfrac{1}{\sqrt2}\cdot\sqrt2\,\cos\!\Big(\alpha-\tfrac{\pi}{4}\Big)=\cos\!\Big(\alpha-\tfrac{\pi}{4}\Big). \]

Evaluate the inner product symbolically using \(\cos\alpha+\sin\alpha=\sqrt2\cos(\alpha-\pi/4)\). B

\[ \alpha=\tfrac{\pi}{4}:\ \ \langle\psi|\phi\rangle=\cos 0=1;\qquad \alpha=\tfrac{3\pi}{4}:\ \ \langle\psi|\phi\rangle=\cos\!\tfrac{\pi}{2}=0. \]

At \(\alpha=\pi/4\) the states coincide (parallel) — equality; at \(3\pi/4\) they are orthogonal — the bound is slack. A

\[ \big|\langle\psi|\phi\rangle\big|=\Big|\cos\!\big(\alpha-\tfrac{\pi}{4}\big)\Big|\le 1=\||\psi\rangle\|\,\||\phi\rangle\| \]

Reading. The transition probability \(|\langle\psi|\phi\rangle|^2=\cos^2(\alpha-\pi/4)\) never exceeds \(1\), saturating only when the states are identical. Cauchy–Schwarz is the guarantee that overlap probabilities stay bounded by unity.

Units check. Normalised state vectors are dimensionless amplitudes; every quantity here is a pure number. Consistent.

\[ \textbf{Triangle inequality for two forces in }\mathbb{R}^3.\quad \vec F_1=(3,4,0)\,\text{N},\ \ \vec F_2=(0,3,4)\,\text{N}. \]

Standard Euclidean inner product \(\langle\vec a,\vec b\rangle=\sum_i a_ib_i\); norm measured in newtons. A

\[ \|\vec F_1\|=\sqrt{9+16+0}=5\ \text{N},\qquad \|\vec F_2\|=\sqrt{0+9+16}=5\ \text{N}. \]

Compute each magnitude before combining. A

\[ \vec F_1+\vec F_2=(3,7,4)\,\text{N},\qquad \|\vec F_1+\vec F_2\|=\sqrt{9+49+16}=\sqrt{74}\approx 8.60\ \text{N}. \]

Resultant magnitude by direct component summation. A

\[ \langle\vec F_1,\vec F_2\rangle=0+12+0=12\ \text{N}^2,\quad |\langle\vec F_1,\vec F_2\rangle|=12\le \|\vec F_1\|\,\|\vec F_2\|=25\ \text{N}^2.\ \checkmark \]

Verify Cauchy–Schwarz numerically; the slack (12 versus 25) reflects a non-zero angle. B

\[ \|\vec F_1+\vec F_2\|=\sqrt{74}\approx 8.60\ \text{N}\ \le\ \|\vec F_1\|+\|\vec F_2\|=10\ \text{N} \]

Reading. The resultant of the two forces (8.60 N) is strictly less than the arithmetic sum of their magnitudes (10 N) because they are not parallel; the deficit measures how much they pull in different directions, with \(\cos\theta=12/25=0.48\), \(\theta\approx61^\circ\).

Units check. All magnitudes and the resultant are in newtons; the inner product and the product \(\|\vec F_1\|\|\vec F_2\|\) are in \(\text{N}^2\). Both inequalities are dimensionally homogeneous.

Problems

For \(\vec a=(1,2,2)\) and \(\vec b=(2,0,-1)\) in \(\mathbb{R}^3\) (dimensionless), verify Cauchy–Schwarz and find the angle between the vectors.
Solution
\(\langle\vec a,\vec b\rangle=1\cdot2+2\cdot0+2\cdot(-1)=0\). Since the inner product is \(0\), \(|\langle\vec a,\vec b\rangle|=0\le\|\vec a\|\,\|\vec b\|=3\cdot\sqrt5\approx6.71\). The vectors are orthogonal, so \(\theta=90^\circ\); the bound is maximally slack.
Two normalised states have overlap \(\langle\psi|\phi\rangle=0.6+0.6i\). Is this consistent with Cauchy–Schwarz, and what is the transition probability?
Solution
\(|\langle\psi|\phi\rangle|=\sqrt{0.6^2+0.6^2}=\sqrt{0.72}\approx0.849\). Both states normalised gives \(\|\psi\|\|\phi\|=1\), and \(0.849\le1\): consistent. The transition probability is \(|\langle\psi|\phi\rangle|^2=0.72\).
In \(L^2[0,1]\) with \(\langle f,g\rangle=\int_0^1 f(x)g(x)\,dx\), take \(f(x)=x\) and \(g(x)=1\). Confirm Cauchy–Schwarz explicitly.
Solution
\(\langle f,g\rangle=\int_0^1 x\,dx=\tfrac12\). \(\|f\|^2=\int_0^1 x^2\,dx=\tfrac13\Rightarrow\|f\|=1/\sqrt3\). \(\|g\|^2=\int_0^1 1\,dx=1\Rightarrow\|g\|=1\). Then \(|\langle f,g\rangle|=\tfrac12\le\|f\|\|g\|=1/\sqrt3\approx0.577\); indeed \(0.5\le0.577\). The slack arises because \(x\) and the constant \(1\) are not proportional.
Show that if \(\|u+v\|=\|u\|+\|v\|\) with \(u,v\neq0\) in a real inner-product space, then \(v=cu\) for some \(c>0\).
Solution
Equality in the triangle inequality forces equality at step 10: \(\mathrm{Re}\langle u,v\rangle=\|u\|\|v\|\), i.e. \(\langle u,v\rangle=\|u\|\|v\|\) in the real case. That is equality in Cauchy–Schwarz, so \(u,v\) are linearly dependent: \(v=cu\). For the sign, \(\langle u,cu\rangle=c\|u\|^2\) must equal \(\|u\|\,\|cu\|=|c|\,\|u\|^2>0\), hence \(c=|c|>0\): parallel and same-direction.
Prove \((a_1b_1+a_2b_2+a_3b_3)^2\le(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\), state the equality condition, and use it to bound \(x+2y+2z\) subject to \(x^2+y^2+z^2=1\).
Solution
This is Cauchy–Schwarz in \(\mathbb{R}^3\) with the Euclidean inner product, \(\vec a=(a_1,a_2,a_3)\), \(\vec b=(b_1,b_2,b_3)\); it follows from the general theorem, with equality iff \(\vec a\parallel\vec b\). Application: put \(\vec a=(x,y,z)\), \(\vec b=(1,2,2)\). Then \(x+2y+2z=\langle\vec a,\vec b\rangle\le\|\vec a\|\|\vec b\|=1\cdot\sqrt{1+4+4}=3\), so \(x+2y+2z\le3\), attained when \((x,y,z)=\tfrac13(1,2,2)\), giving \(\tfrac13(1+4+4)=3\). The minimum is \(-3\) at the antiparallel point.