If $\|\cdot \|$ is the norm induced by the inner product $\langle,\rangle$, how to prove the following interesting inequality? $$\langle x,y\rangle(\|x\|+\|y\|) \leq\|x+y\|\,\|x\|\,\|y\|$$
This is a exercise in my textbook which is available only in portuguese called "Topologia e Análise no Espaço $\mathbb{R}^n$". The above inequality is obvious when $\langle x,y\rangle \leq 0$, but I don't know how to proceed to prove the other case.
Best Answer
If $\langle x,y \rangle \leq 0$, then the result clearly holds. Otherwise, assume $\langle x,y \rangle >0$.
$$\frac{{\| x+y \|}^2}{(\| x \| + \| y \| )^2} = \frac{{\| x \|}^2 + {\| y \|}^2 +2\langle x,y \rangle}{{\| x \|}^2 + {\| y \|}^2 + 2 \| x \| \| y \|} \geq \frac{2 \langle x,y \rangle }{2 \| x \| \| y \|} = \frac{\langle x,y \rangle }{\| x \| \| y \|} \geq \frac{{ \langle x,y \rangle }^2}{{\| x \|}^2 {\| y \|}^2}$$ where the inequality comes from $\frac{| \langle x,y \rangle |}{\| x \| \| y \|} \leq 1$ (the Cauchy-Schwarz ineqaulity).
Take the square root of both sides to get the result: $$ 0 \leq \frac{\langle x,y \rangle}{\| x \| \| y \|} \leq \frac{\| x+y \|}{\| x \| + \| y \|} \leq 1.$$