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For a matrix with nonnegative entries, row sums at most $R$ and column sums at most $C$ imply an ${\ell }^2$ operator norm at most $\sqrt{RC}$. Indeed the Cauchy-Schwarz inequality gives $|{\sum }_j{A}_{ij}{x}_j{|}^2\le ({\sum }_j{A}_{ij}){\sum }_j{A}_{ij}|x_j{|}^2$. Summing over $i$ bounds the result by $RC{\sum }_j|x_j{|}^2$. Integral kernels and positive weight functions give analogous forms. The symmetric case has equal row and column bounds.

For a nonnegative random variable $N$ with positive expected value and finite second moment, write $N=N{1}_{\{N>0\}}$. The Cauchy-Schwarz inequality giveswhich proves the bound in second moment method. It is the zero-threshold case of the Paley-Zygmund inequality. Equality occurs when $N$ is a fixed positive value on the event $N>0$ and zero elsewhere, almost surely. For $N=10B$ with $B$ a Bernoulli random variable of parameter $1/10$, $(\mathbb{E}N{)}^2/\mathbb{E}{N}^2=1/10$ equals the exact positivity probability.

The variance identity $\mathbb{E}{N}^2=(\mathbb{E}N{)}^2+\mathrm{Var}(N)$ turns the lower bound into a statement about variance relative to the squared mean. A small ratio forces positivity to be likely. A positive expected value alone does not do so: keeping $\mathbb{E}N=1$ while taking $N=1/q$ with probability $q$ and zero otherwise makes the positivity probability as small as any chosen $q>0$.

Put $L=\sqrt{D}$, $h(t)={\prod }_{p\mid t}(g(p{)}^{-1}-1)$ and $y_t={\sum }_{\begin{array}{c}m\le L\\ t\mid m\end{array}}g(m){\rho }_m$, where $g$ is a multiplicative arithmetic function supported on squarefree integers. For squarefree integers $m,n$,Thus the quadratic form is $\Sigma ={\sum }_{t\le L}h(t)y_t^2$. Möbius inversion gives $g(d){\rho }_d={\sum }_{d\mid t\le L}\mu (t/d)y_t$. On the permitted squarefree integers, the constraint ${\rho }_1=1$ becomes ${\sum }_t\mu (t)y_t=1$. The Cauchy-Schwarz inequality yields $\Sigma \ge 1/J$, where $J={\sum }_{t}1/h(t)$ over those integers; equality holds at $y_t=\mu (t)/(Jh(t))$.

With the closed-boundary assumptions of the resistive magnetic-helicity balance, define the nonnegative resistive magnetic energy loss $D_B={\eta }_m{\int }_V|\nabla \times \mathbf{B}{|}^{2}dV/{\mu }_0$. The Cauchy-Schwarz inequality gives the displayed bound. Consequently, in a family of fields with bounded $E_B,D_B$ as ${\eta }_m\to 0$, resistive magnetic helicity loss tends to zero even when $D_B$ remains finite. This is an absolute conservation estimate; it is not a universal comparison of fractional decay times. For a single Beltrami field with $\nabla \times \mathbf{B}=k\mathbf{B}$ and no motion, both $E_B$ and $H$ have the same fractional decay rate $2{\eta }_m{k}^2$.

For $(u,s),(v,t)$ in the second-order cone, the Cauchy-Schwarz inequality gives $u\cdot v+st\ge -\Vert u\Vert \Vert v\Vert +st\ge 0$. Thus the cone is contained in its dual cone. Conversely, if $s<\Vert u\Vert$ and $u\ne 0$, pairing $(u,s)$ with $(-u/\Vert u\Vert ,1)$ is negative. If $u=0$ and $s<0$, use $(0,1)$. Every point outside the cone therefore lies outside its dual, proving that it is a self-dual cone.

The spatial Galerkin method for the heat equation gives a mass matrix $M$ and a diffusion stiffness matrix $S$. With zero forcing, $\frac{d}{dt}(c^{T}Mc)=-2c^{T}Sc\le 0$. The quadratic form $c^{T}Mc$ is the squared L2 norm of the represented finite-element function, proving an energy contraction. For nonzero forcing, the corresponding identity has the additional term $2c^{T}F$, which can be estimated using the Cauchy-Schwarz inequality.

For positive traffic rates in single-server colonies, a total service budget $F>{\sum }_j{\alpha }_j$ minimizes the stationary mean population when the spare rates ${\varphi }_j-{\alpha }_j$ are proportional to $\sqrt{{\alpha }_j}$. The Cauchy-Schwarz inequality proves the optimum directly.

For $y(0)=y(1)=0$, write $A={\int }_0^{1}ydx$ and $w=x(1-x)$. Integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, while ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality therefore gives the displayed sharp bound, with equality exactly when $y=Kw$. A prescribed positive budget $B={\int }_0^1(y+y^{\prime 2})dx$ consequently givesThis proves the unique global optimum, rather than only stationarity from a Lagrange multiplier.

For $x\in {\mathbb{R}}^n$, Cauchy-Schwarz inequality applied to $(|x_i|)$ and $(1)$ gives the upper bound. Equal absolute coordinate values attain its factor $\sqrt{n}$. Expanding $({\sum }_i|x_i|{)}^2$ gives the lower bound, and a vector with only one nonzero coordinate attains its factor one. Thus both constants are optimal. These Lp norms define the same finite-dimensional topology despite having different unit balls.

Assume a positive definite covariance matrix and nonzero excess-mean vector $d=\mu -r1$. The Cauchy-Schwarz inequality gives $|d^{T}x|\le \sqrt{d^T{\Sigma }^{-1}d}\sqrt{x^T\Sigma x}$, with the positive-premium efficient direction $x=cz$, $c>0$. If $s=1^{T}z\ne 0$, its normalized risky portfolio has Sharpe ratio $\mathrm{sgn}(s)\sqrt{d^T{\Sigma }^{-1}d}$. Thus negative normalization reverses the efficient orientation. If $s=0$, the efficient risky direction is self-funded and cannot be normalized to a unit risky budget.

A skew-coupled elliptic system has zeroth-order couplings with opposite signs. For the displayed example with homogeneous Dirichlet boundary conditions, its weak formulation on $V=H_0^1(\Omega )\times H_0^1(\Omega )$ has the bilinear formThe coupling cancels for equal arguments, so $b((p,q),(p,q))=\Vert \nabla p{\Vert }_0^2+\Vert \nabla q{\Vert }_0^2$. On a domain satisfying the Poincaré inequality, this is a coercive bilinear form. With the product first-order Sobolev norm, the Cauchy-Schwarz inequality applied to $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$ gives $|b(P,T)|\le \Vert P{\Vert }_V\Vert T{\Vert }_V$. The Lax-Milgram theorem therefore applies even though the bilinear form is not symmetric. Eliminating $w=-\Delta u$ recovers ${\Delta }^{2}u+u=f$, involving the biharmonic operator.

The continuous dual space of $H^s({\mathbb{R}}^d)$ is $H^{-s}({\mathbb{R}}^d)$ under the extension of the L2 inner product. The Fourier transform definition $\Vert f{\Vert }_{H^s}^2=\int (1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi$ proves boundedness of the pairing by the Cauchy-Schwarz inequality, and the Riesz representation theorem proves that every continuous functional has this form. Consequently a bounded linear operator $K:H^r\to H^{r+a}$ has an adjoint operator, relative to this pairing, from $H^{-r-a}$ to $H^{-r}$. For an operator that smooths by two orders for every $r$, this yields $K^{\ast }:H^2\to H^4$ by taking $r=-4$.

Let $H=M-\delta J$ be the centered biadjacency matrix of an equal-part biregular graph, and put $s=\Vert H{\Vert }_{\mathrm{op}}$. For subsets $A,B$ of the two parts, their edge discrepancy is $1_A^{\mathsf{T}}H{1}_B$. The Cauchy-Schwarz inequality proves the displayed bound. If the part size is $n$, centering the two indicator vectors improves it to $sn\sqrt{\alpha (1-\alpha )\beta (1-\beta )}$. The centered four-cycle identity for a biregular graph shows that $t_4(G)\le {\delta }^4(1+c^4)$ implies $s\le c\delta n$ for $c\ge 0$.

For equally sized Hermitian matrices and a real test function with Lipschitz bound one, the difference of its averages under their empirical spectral measures is at most $N^{-1/2}$ times the Frobenius norm of their difference. Pair the increasing eigenvalues, apply the triangle inequality and Cauchy-Schwarz inequality, then use the Hoffman–Wielandt inequality.

Positivity makes $A-z$ bounded below with dense closed range for every $z<0$, excluding negative spectral values. Conversely if the spectrum is nonnegative, $R=(I+A{)}^{-1}$ has spectrum in $[0,1]$ by the resolvent spectral mapping identity. The numerical-range endpoint approximate-eigenvalue theorem gives $0\le R\le I$, and positive-form Cauchy-Schwarz inequality gives $\Vert Rf{\Vert }^2\le ⟨Rf,f⟩$. For $u=Rf$, $⟨Au,u⟩=⟨Rf,f⟩-\Vert Rf{\Vert }^2\ge 0$.

In three dimensions let $v\in H_0^1(B_R)$ solve $-\Delta v=f$, where $f\in L^{6/5}(B_R)$ is supported in $B_a$ and $0<a<R$. Write $M=\int f$ and $\bar{v}(s)=(4\pi {)}^{-1}{\int }_{S^2}v(s\omega )d\omega$. Outside $B_a$, the weak solution is a harmonic function. The weak formulation against radial test functions gives $-4\pi s^2{\bar{v}}^{\prime }(s)=M$ for $a<s<R$. The zero Sobolev trace gives $\bar{v}(R)=0$; integration proves the displayed formula, without assuming the source is a radial function.

Cauchy-Schwarz inequality on spheres gives $\Vert v{\Vert }_2^2\ge M^2(4\pi {)}^{-1}{\int }_a^R(1-s/R{)}^{2}ds$. For fixed nonzero $M$ and fixed $a$, this grows linearly in $R$. Such solutions on expanding balls cannot remain bounded in the full $H^1({\mathbb{R}}^3)$ norm after zero extension of H01.

Under the hypotheses of Bishop volume comparison with a positive-curvature model, the Bonnet-Myers theorem gives diameter at most $D$. If total Riemannian volume equals $v_k(D)$, every comparison ball ratio is one. In a small normal ball the pointwise radial density bound therefore becomes equality. Writing $h={\partial }_r\log J$, the traced radial Riccati equation for distance spheres and Cauchy-Schwarz inequality force the radial operator to equal $(s_k^{\prime }/s_k)I$. The angular metric is consequently $s_k(r{)}^2{g}_{S^{n-1}}$, so the metric is locally round. The classification of complete positive constant-curvature manifolds gives a quotient $S_k^n/\Gamma$, whose volume is $v_k(D)/|\Gamma |$. Equality forces $\Gamma$ trivial.

For mean-zero $f$, apply the mean ergodic theorem on an ergodic product system to $f\otimes \overline{f}$, paired with $g\otimes \overline{g}$. Its correlation is $|⟨U^{n}f,g⟩{|}^2$. The Cauchy-Schwarz inequality then gives averaged absolute correlation decay, the definition of weak mixing.

Use the unnormalized Fourier transform and $|f|\le 1$. Normalize each derivative correlation by $N$ and align its phase. Their absolute sum is at least $\alpha N$. One Cauchy-Schwarz inequality, expansion by shift/frequency differences, and a second Cauchy-Schwarz inequality bound its fourth power by $N$ times the additive energy of a frequency graph. Parseval identity on a finite group supplies the factor $N$; exact equality of both shift sums and frequency sums is retained.

Every finite point set in the unit square admits a cyclic visiting order with ${\sum }_i\Vert x_{i+1}-x_i{\Vert }^2\le 4$. Split the square into two right triangles, apply the quadratic path bound in a right triangle, and remove auxiliary corners using the law of cosines. The Cauchy-Schwarz inequality then gives $\mathrm{TS}(X)\le 2\sqrt{|X|}$. The tour minimizing ordinary length need not minimize the sum of squared lengths.