inequality in Hilbert space

Question

Let $X$ be a real Hilbert space.Let $x,y \in X$ such that $\langle x,y\rangle >0$. If $\alpha \geq 1$.

I want to prove that $\Vert \alpha x-y \Vert \leq \Vert x-y \Vert$

Answer 1

Inequality is wrong : for $x=y \ne 0$ and $\alpha = 0$ you get $||\alpha x-y|| = ||y|| > 0 = ||x-y||$

Answer 2

In $\Bbb R$, let $x = 1, y = 1, \alpha = 0.5$. The right hand side is $0$, the left hand side is $0.5$. The thing you're trying to prove is false.

Answer 3

The inequality doesn't hold true in general. What you can show is this: $$\|\alpha x+y \|^2=\langle \alpha x+y, \alpha x+y\rangle=|\alpha|^2\|x\|^2+2\alpha\langle x,y\rangle+\|y\|^2 \ge ||x+y||^2$$