# Inequality involving norm and inner product

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.

• Squaring cannot work, since for $y=-x$ the LHS is negative and the RHS is $0$. Sep 3, 2012 at 15:46
• If the inner product is negative then the inequality holds. Otherwise I square both sides.
– p.t.
Sep 3, 2012 at 15:55
• 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.... Sep 3, 2012 at 16:03
• 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.
– Theo
Sep 3, 2012 at 16:04
• @N.S.: I don't understand your objection. Sep 3, 2012 at 17:03

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.

• Many thanks! I kept using C-S to completely cancel the denominator; that's why it doesn't work.
– p.t.
Sep 3, 2012 at 23:48

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.

• 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.
– Theo
Sep 3, 2012 at 17:57
• @Theo: Thanks, that's much simpler than my brute-force approach. Sep 4, 2012 at 3:05