Inequality involving inner product and norm

If $\langle x,y \rangle \leq 0$, then the result clearly holds. Otherwise, assume $\langle x,y \rangle >0$.

$$\frac{{\| x+y \|}^2}{(\| x \| + \| y \| )^2} = \frac{{\| x \|}^2 + {\| y \|}^2 +2\langle x,y \rangle}{{\| x \|}^2 + {\| y \|}^2 + 2 \| x \| \| y \|} \geq \frac{2 \langle x,y \rangle }{2 \| x \| \| y \|} = \frac{\langle x,y \rangle }{\| x \| \| y \|} \geq \frac{{ \langle x,y \rangle }^2}{{\| x \|}^2 {\| y \|}^2}$$ where the inequality comes from $\frac{| \langle x,y \rangle |}{\| x \| \| y \|} \leq 1$ (the Cauchy-Schwarz ineqaulity).

Take the square root of both sides to get the result: $$ 0 \leq \frac{\langle x,y \rangle}{\| x \| \| y \|} \leq \frac{\| x+y \|}{\| x \| + \| y \|} \leq 1.$$