3 Simple Proofs of the Cauchy-Schwarz Inequality

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Theorem. (Cauchy-Schwarz Inequality)
Let $X$ be an inner product space with a given inner product $ \langle \cdot , \cdot \rangle : X \times X \to \mathbb F ( \mathbb R \; \text{or} \; \mathbb C ) $. Then $$ | \langle x,y \rangle |^2 \leqslant \langle x,x \rangle \langle y,y \rangle $$ for any $x,y \in X$.

Proof (1)
If $y = 0$ then it’s trivial. Let $y \neq 0$ and define $z := x – \frac{ \langle x,y \rangle}{\langle y,y \rangle} y$ so that $\langle z,y \rangle = 0$. Then we have
\begin{eqnarray}
\langle x,x \rangle &=& \left \langle \frac{\langle x,y \rangle}{\langle y,y \rangle} y + z , \frac{\langle x,y \rangle}{\langle y,y \rangle} y + z \right \rangle \\ &=& \frac{| \langle x,y \rangle |^2}{( \langle y,y \rangle )^2} \langle y,y \rangle + \langle z,z \rangle \\ & \geqslant &  \frac{| \langle x,y \rangle |^2}{\langle y,y \rangle}
\end{eqnarray}
which proves the Theorem.

Proof (2)
If $ | \langle x,y \rangle | = 0$ then it’s trivial. Let $| \langle x,y \rangle | \neq  0$ and let $ \lambda \in \mathbb R$. Then
\begin{eqnarray} 0 & \leqslant & \langle x + \lambda \langle x,y \rangle y ,  x + \lambda \langle x,y \rangle y \rangle \\ &=& | \langle x,y \rangle |^2 \langle y,y \rangle \lambda^2 + 2 | \langle x,y \rangle |^2 \lambda + \langle x,x \rangle .
\end{eqnarray}
Because all the coefficients are real, we have the discriminant $$ D/4 = | \langle x,y \rangle |^4 – | \langle x,y \rangle |^2 \langle x,x \rangle \langle y,y \rangle \leqslant 0 $$ which proves the Theorem.

Proof (3)
Let $\alpha \in \mathbb C$. Then
\begin{eqnarray}
0 & \leqslant & \langle x+ \alpha y , x + \alpha y \rangle \\ &=& \langle x,x \rangle + \overline{\alpha} \langle x,y \rangle + \alpha \langle y,x \rangle + | \alpha |^2 \langle y,y \rangle \\ &=& \begin{bmatrix} 1 & \alpha \end{bmatrix} \begin{bmatrix}  \langle x,x \rangle & \langle x,y \rangle \\  \langle y,x \rangle & \langle y,y \rangle \end{bmatrix} \begin{bmatrix}  1\\ \overline{\alpha} \end{bmatrix} \end{eqnarray}
Define $$ A = \begin{bmatrix}  \langle x,x \rangle & \langle x,y \rangle \\  \langle y,x \rangle & \langle y,y \rangle \end{bmatrix} $$
 
then $A$ is Hermitian. If $z = [ \beta \;\;\;\; \overline{\alpha}]^T $ for $ \beta \neq 0$, $\overline{z}^T A z \geqslant 0 $ by considering $ z / \beta = [1 \;\;\;\; \overline{\alpha} / \beta ]^T $. If $ \beta = 0$, $\overline{z}^T A z = | \alpha |^2 \langle y,y \rangle \geqslant 0$. Thus $A$ is positive semidefinite, so has non-negative determinant $$ 0 \leqslant \det A = \langle x,x \rangle \langle y,y \rangle – | \langle x,y \rangle |^2 $$
 
which proves the Theorem.

References.
[1] http://ckrao.wordpress.com/2011/02/18/two-interesting-proofs-of-the-cauchy-schwarz-inequality-complex-case