# For $||x|| \leq ||y||$ is it true that $||x||^4 + ||y||^4 - 2\langle x, y \rangle ^2 \leq 3 ||y||^2 ||x-y||^2$?

I have recently come across this seemingly simple Hilbert space inequality

$$||x||^4 + ||y||^4 - 2\langle x, y \rangle ^2 \leq 3 ||y||^2 ||x-y||^2$$

for $||x|| \leq ||y||$, where $<\cdot, \cdot>$ denotes the inner product and $||\cdot||$ the corresponding norm. I have been trying to prove it but something always seems to be off. If the inequality holds, could you please give me a pointer on how to prove it?

• I have rapidly shown a nearby inequality using Cauchy Schwartz and $(x+a)^2\geqslant\frac{x^2}{2}-a^2$: $\|y\|^4-2<x,y>^2 \leqslant 2\|y\|^2\|x-y\|^2$ Maybe it can help? Jul 19, 2018 at 8:26
• @BenoitGaudeul Would you like to post a full solution as an answer so I can mark it as correct? Jul 19, 2018 at 8:42
• A full solution of this lesser inequality? because for the full problem I'm not so sure. Jul 19, 2018 at 8:55
• Of the original inequality starting from the easily proved lesser inequality $(x+a)^2\geqslant\frac{x^2}{2}-a^2$ Jul 19, 2018 at 8:55

Let's consider our Hilbert space to be $\mathbb{R}$ and let $y=1$, $x=1-h$
$(1-h)^4+1-2(1-h)^2=4h^2-4h^3+h^4$
The right-hand side is $3h^2$. For $h$ smaller than $2-\sqrt3$, the inequality doesn't hold.