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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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    Squaring cannot work, since for $y=-x$ the LHS is negative and the RHS is $0$.2012-09-03
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    If the inner product is negative then the inequality holds. Otherwise I square both sides.2012-09-03
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    True, but once you square it the sign becomes irrelevant in the inner product. It is pretty improbable to have an inequality consisting of even functions, which is true only when an expression is positive....2012-09-03
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    This is equivalent with the following: If a, b, c are the sides of an obtuse triangle (with c the longest side), then $\cos(x)\leq\frac{c}{a+b}$. where $x$ is $\pi$-(obtuse angle). It should work with a geometric argument.2012-09-03
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    I get the interpretation in geometry but a proof using only linear algebra is much better for me. Anyway, I'll try out your suggestion.2012-09-03
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    @N.S.: I don't understand your objection.2012-09-03
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    Guys, you are missing the point. My point is that we know that for negative $A$ sometimes $A^2 \leq B^2$ doesn't work. Yes, IF you can prove that $A^2 \leq B^2$ that you get the desired inequality... BUT I really fail to see why would an inequality of the type $A^2 \leq B^2$ always hold for positive $A$ and fail sometimes for negative $A$...2012-09-03
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    So, my point is that the approach to square it doesn't seem the best approach.... Is not something with the problem, is just that the method mentioned in the post seems bound to fail...2012-09-03
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    @anon If you prove that $A^2 \leq B^2$ than your approach works... But, if you KNOW that $A^2 \geq B^2$ for many negative $A$,would you try to use your approach?2012-09-03
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    @N.S. Right... don't mind me...2012-09-03
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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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    You don't need any calculus to verify the inequality. If you expand and move everything to the right, you get $(1-\alpha^2)+r(1-\alpha^2)+2r(1-\alpha)\geq 0$, which is trivially true, given that $r\geq 0$ and $0\leq \alpha\leq 1$. Thus, it becomes a purely algebraic solution.2012-09-03
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    @Theo: Thanks, that's much simpler than my brute-force approach.2012-09-04