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I am stuck proving this trivial inequality: on a real inner product space,

$(||x||+||y||)\frac{\langle x,y\rangle}{||x|| \cdot ||y||}\leq||x+y||$

I have tried to square both sides and use the Cauchy Schwarz inequality to get to $||x||\cdot||y||\leq\langle x,y\rangle$, which is obviously incorrect.

Any help is much appreciated.

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    I should had made my first post more clear. Also I should point out that replacing $A$ by $-A$ changes the expression in $B$, so squaring is not necessarily bound to fail, but the point I tried to make with my first post is that because of that example, my intuition tells me that squaring both sides should not be helpfull. Of course, my intuition fails me sometimes ;)2012-09-03

2 Answers 2

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When $\langle x,y \rangle \leq 0$ left side is negative (or zero), so the claim is trivial.

When $\langle x,y \rangle > 0$, then we can square both sides and $ (\|x\|+\|y\|)^2\frac{{\left\langle x,y \right\rangle} \overbrace{{\left\langle x,y \right\rangle}}^{\text{C-S this}} }{\| x \|^2 \| y \|^2} \leq \left(\|x\|^2+2\|x\|\|y\|+\|y\|^2\right)\frac{\left\langle x,y \right\rangle}{\|x\|\|y\|} \text{.} $ RHS becomes $ \frac{\|x\|}{\|y\|}\overbrace{\left\langle x,y \right\rangle}^\text{C-S this}+2\left\langle x,y \right\rangle+\frac{\|y\|}{\|x\|}\overbrace{\left\langle x,y \right\rangle}^\text{C-S this} \leq \|x\|^2 + 2\left\langle x,y \right\rangle + \|y\|^2 = \| x+y \|^2 $ which gives the claim.

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    Many thanks! I kept using C-S to completely cancel the denominator; that's why it doesn't work.2012-09-03
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Since the inequality is invariant under the scaling $(x,y)\mapsto (tx,ty)$, we can assume $\|x\|=1$. Set $r=\|y\|$ and $\alpha = \langle x,y \rangle / r \in [1,1]$. We now have the inequality $(1+r)\alpha \le \sqrt{1 + r^2 + 2 r \alpha}, \quad r \ge 0,\, \alpha \in [-1,1]$ which should be a simple calculus exercise to verify.

It is trivial when $\alpha < 0$, so let us assume $\alpha \ge 0$. In this case we can square both sides to get the equivalent inequality $(1+r)^2 \alpha^2 \le 1+r^2+2r\alpha.$ So if we set $F(r,\alpha) = 1+r^2+2r\alpha-(1+r)^2\alpha^2, \quad r \ge 0, \,\alpha \in [0,1]$ we can see $F \ge 0$ for $r = 0$, $r \to +\infty$, $\alpha = 0$ and $\alpha = 1$. It remains to look for critical points of $F$ inside the region $r > 0$, $\alpha \in (0,1)$. However, differentiating with respect to $r$, we have $F_r(r,\alpha) = 2r + 2\alpha - 2\alpha^2(1+r) = 2r(1-\alpha^2) + 2\alpha(1-\alpha)$ which is strictly positive on $r > 0$, $\alpha \in (0,1)$. So there are no critical points, and we have $F \ge 0$ on the entire region.

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    @Theo: Thanks, that's much simpler than my brute-force approach.2012-09-04