How to show that $\|a+b+c\|^2\leq 3\|a\|^2+3\|b\|^2+3\|c\|^2$ Show that $$\|a+b+c\|^2\leq 3\|a\|^2+3\|b\|^2+3\|c\|^2$$
where $a,b,c$ are in some Hilbert space $(H,\langle\cdot,\cdot \rangle)$?
I see that we have $$\|a+b\|^2\leq2\|a\|^2 +2 \|b\|^2$$ due to the parallelogram law. What about the first inequality?

You can expand the LHS and use $\left<x,y\right>+\left<y,x\right>\leq ||x||^2+||y||^2$ 3 times. This latter inequality follows by expanding the non-negative entity $\left<x-y,x-y\right>$.
