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.
 A: 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.
A: 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.
