8
$\begingroup$

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.

$\endgroup$
0

3 Answers 3

4
$\begingroup$

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: $$ 0 \leq \frac{\langle x,y \rangle}{\| x \| \| y \|} \leq \frac{\| x+y \|}{\| x \| + \| y \|} \leq 1.$$

$\endgroup$
2
  • $\begingroup$ But if we consider a non zero vector $x$ and take the inner product with $y=-x$ the inequality fails to be true with the absolute value of the inner product, instead of only the inner product. $\endgroup$
    – Victor
    Feb 16, 2019 at 21:21
  • 2
    $\begingroup$ You are very right and I've editted my answer. I think this is now a good, and quite straightforward answer. It gives a clear idea that the Cauchy-Schwarz inequality is further from being an equality than the triangle inequality, which is interesting. $\endgroup$ Feb 16, 2019 at 21:40
2
$\begingroup$

(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 $\langle{x,y}\rangle \geqslant 0$.)

$\endgroup$
3
  • $\begingroup$ Actually, I loved your solution, it's very clear :) $\endgroup$
    – Victor
    Feb 16, 2019 at 21:07
  • 1
    $\begingroup$ Thank you. But seriously, if someone does post a more insightful solution, you can "unaccept" my answer and accept the other one - I won't be offended! $\endgroup$ Feb 16, 2019 at 21:14
  • 1
    $\begingroup$ Indeed, Lemniscate's answer is much nicer, and probably actually memorable - whereas I'll have forgotten how this calculation went by tomorrow. :) $\endgroup$ Feb 16, 2019 at 22:36
1
$\begingroup$

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?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .