Proof of Cauchy-Schwarz Inequality in a Complex Inner Product Space

Conditions 1-4 below are those of Complex Inner Product Space. We use the properties of conjugation and of the modulus recorded in Properties of Complex Conjugation and Modulus, the elementary inner-product identities of Elementary Properties of a Complex Inner Product, and the vector-space identities of Elementary Identities in a Vector Space. We also use the following fact about ordered fields: if $p\le q$ and $0\le r$ are real numbers, then $pr\le qr$, since $0\le(q+(-p))r=qr+(-(pr))$ by the second order-compatibility condition, and adding $pr$ gives the assertion.

Write $s=\langle v,v\rangle$ and $c=\langle v,u\rangle$. By condition 4, $s$ is a real number with $0\le s$.

Case 1: $s=0$. Then $v=0_{V}$ by claim 4 of Elementary Properties of a Complex Inner Product, so $\langle u,v\rangle=0$ by claim 3 there. Both sides of the asserted inequality are then $0$: the left side because $|0|=0$ by claim 3 of Properties of Complex Conjugation and Modulus, the right side because $\langle u,u\rangle\cdot0=0$.

Case 2: $s\neq0$. Then $s$ has a multiplicative inverse $s^{-1}$, which is again real by claim 1 of Canonical Form and Arithmetic of Complex Numbers. Put $t=s^{-1}c$ and $x=u+\bigl((-t)v\bigr)$, where $(-t)v=-(tv)$ by claim 5 of Elementary Identities in a Vector Space together with condition 5 of Vector Space over a Field.

Expanding $\langle x,x\rangle$ by additivity in each argument (condition 2 and claim 1 of Elementary Properties of a Complex Inner Product), homogeneity in the second argument (condition 3) and conjugate homogeneity in the first (claim 2 there),

Now $\langle u,v\rangle=\overline{\langle v,u\rangle}=\overline{c}$ by condition 1, and $\overline{t}=\overline{s^{-1}c}=s^{-1}\overline{c}$, because conjugation is multiplicative and fixes the real number $s^{-1}$ by claim 1 of Properties of Complex Conjugation and Modulus. Using $z\overline{z}=|z|^{2}$ (claim 3 there) and $\overline{(-t)}(-t)=\overline{t}t$,

Adding the three contributions,

and this is a real number with $0\le\langle x,x\rangle$ by condition 4. Adding $s^{-1}|c|^{2}$ to both sides of $0\le\langle u,u\rangle-s^{-1}|c|^{2}$ gives

Multiplying by the nonnegative real number $s$, by the ordered-field fact stated above, and using $(s^{-1}|c|^{2})s=|c|^{2}$, we obtain $|c|^{2}\le\langle u,u\rangle\,s$.

Finally $|c|=|\langle v,u\rangle|=|\overline{\langle u,v\rangle}|=|\langle u,v\rangle|$ by condition 1 and claim 3 of Properties of Complex Conjugation and Modulus. Hence

which is the asserted inequality.