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.
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We need to prove that $$\|x+y\|\left(\|x\|\|y\|-\langle x,y\rangle\right)\geq\langle x,y\rangle\left(\|x\|+\|y\|-\|x+y\|\right)$$ or $$\left(\|x\|+\|y\|+\|x+y\|\right)\|x+y\|\left(\|x\|\|y\|-\langle x,y\rangle\right)\geq2\langle x,y\rangle\left(\|x\|\|y\|-\langle x,y\rangle\right),$$ for which it's enough to prove that $$\left(\|x\|+\|y\|+\|x+y\|\right)\|x+y\|\geq2\langle x,y\rangle.$$ Now, by the triangle inequality $$\left(\|x\|+\|y\|+\|x+y\|\right)\|x+y\|\geq\left(2\|x+y\|\right)\|x+y\|=2\langle x+y,x+y\rangle.$$ Can you end it now?
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(I don't suggest you do it this way, because it's just a mindless calculation. I hope someone posts a more insightful solution. I have no idea what's going on here.)
By the Cauchy-Schwarz inequality, $$ \langle{x,y}\rangle \leqslant \|x\|\|y\|. $$
Squaring, and multiplying by $\|x\|^2 + \|y\|^2$, $$ \langle{x,y}\rangle^2\left(\|x\|^2 + \|y\|^2\right) \leqslant \|x\|^2\|y\|^2\left(\|x\|^2 + \|y\|^2\right). $$
Alternatively, multiplying by $2\langle{x,y}\rangle\|x\|\|y\|$ (without squaring it first), $$ 2\langle{x,y}\rangle^2\|x\|\|y\| \leqslant 2\langle{x,y}\rangle\|x\|^2\|y\|^2. $$
Adding these two inequalities, \begin{align*} \langle{x,y}\rangle^2\left(\|x\|^2 + \|y\|^2 + 2\|x\|\|y\|\right) & \leqslant \|x\|^2\|y\|^2\left(\|x\|^2 + \|y\|^2 + 2\langle{x,y}\rangle\right) \\ & = \|x\|^2\|y\|^2\|x + y\|^2, \end{align*} and now it's just a matter of taking the square roots of both sides.
(I've assumed throughout that $\langle{x,y}\rangle \geqslant 0$.)
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: $$ 0 \leq \frac{\langle x,y \rangle}{\| x \| \| y \|} \leq \frac{\| x+y \|}{\| x \| + \| y \|} \leq 1.$$