How to show that $\|a+b+c\|^2\leq 3\|a\|^2+3\|b\|^2+3\|c\|^2$

Question

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?

Answer 1

You can prove that $2 | \langle a, b \rangle | \leq \| a \|^2 + \| b \|^2$ as I believe you have proven to attain $\| a + b \|^2 \leq 2\| a \|^2 + 2\| b \|^2$.

Then you just expand:

$\| a + b + c \|^2 = \langle a, a \rangle + \langle b, b \rangle + \langle c, c \rangle + 2\langle a, b \rangle + 2\langle a, c \rangle + 2\langle b, c \rangle \leq \| a \|^2 + \| b \|^2 + \| c \|^2 + \| a \|^2 + \| b \|^2 + \| a \|^2 + \| c \|^2 + \| b \|^2 + \| c \|^2 = 3 ( \| a \|^2 + \| b \|^2 + \| c \|^2)$

Sometimes, you just have to be brave to dive into the ugly formulas to simplify...

Answer 2

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 $\left<x-y,x-y\right>$.