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.

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.
