Inequality involving inner product and norm 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.
 A: 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: $$ 0 \leq \frac{\langle x,y \rangle}{\| x \| \| y \|} \leq \frac{\| x+y \|}{\| x \| + \| y \|} \leq 1.$$
A: (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 $\langle{x,y}\rangle \geqslant 0$.)
A: 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? 
