Proof of The Cauchy–Schwarz Inequality in a Real Inner Product Space

Throughout, $\|x\|\ge0$ and $\|x\|^{2}=\langle x,x\rangle$ for $x\in V$, since $\|x\|$ is the nonnegative real number whose square is $\langle x,x\rangle$ (Real Inner Product Spaces (vii), The k-th Root and the Square Root of a Nonnegative Real Number (i)).

Expanding a square. For $w,x,y\in V$ and $t\in\mathbb{R}$, symmetry together with additivity and homogeneity in the first argument (Real Inner Product Spaces (i), Real Inner Product Spaces (ii), Real Inner Product Spaces (iii)) gives linearity in the second argument:

$$\langle w,x+ty\rangle=\langle x+ty,w\rangle=\langle x,w\rangle+t\,\langle y,w\rangle=\langle w,x\rangle+t\,\langle w,y\rangle .$$

Expanding first in the first argument and then in the second, and using symmetry once more, we obtain for every $t\in\mathbb{R}$

$$\langle u+tv,u+tv\rangle=\langle u,u+tv\rangle+t\,\langle v,u+tv\rangle=\langle u,u\rangle+2t\,\langle u,v\rangle+t^{2}\langle v,v\rangle .$$

The squared inequality. We show

$$\langle u,v\rangle^{2}\le\|u\|^{2}\,\|v\|^{2}.$$

If $v=0_V$, then $v=0\,0_V$ (Elementary Identities in a Vector Space (ii)), so $\langle v,u\rangle=0\,\langle 0_V,u\rangle=0$ by homogeneity and $\langle u,v\rangle=0$ by symmetry; the left side is $0$ and the right side is a product of squares, so the inequality holds. If $v\ne0_V$, then $\langle v,v\rangle>0$ by Real Inner Product Spaces (iv). Taking $t=-\langle u,v\rangle/\langle v,v\rangle$ in the expansion and using positive definiteness for $u+tv$,

$$0\le\langle u+tv,u+tv\rangle=\langle u,u\rangle-2\,\frac{\langle u,v\rangle^{2}}{\langle v,v\rangle}+\frac{\langle u,v\rangle^{2}}{\langle v,v\rangle}=\langle u,u\rangle-\frac{\langle u,v\rangle^{2}}{\langle v,v\rangle}.$$

Multiplying by $\langle v,v\rangle>0$ gives $\langle u,v\rangle^{2}\le\langle u,u\rangle\langle v,v\rangle=\|u\|^{2}\|v\|^{2}$.

Taking square roots. Both sides of the squared inequality are nonnegative. The square root is monotone, multiplicative, and satisfies $\sqrt{z^{2}}=|z|$ for $z\in\mathbb{R}$ and $\sqrt{a^{2}}=a$ for $a\ge0$ (Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (iv), Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (iii), Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (ii), Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (i)). Therefore

$$|\langle u,v\rangle|=\sqrt{\langle u,v\rangle^{2}}\le\sqrt{\|u\|^{2}\,\|v\|^{2}}=\sqrt{\|u\|^{2}}\,\sqrt{\|v\|^{2}}=\|u\|\,\|v\|,$$

which is (i).