Prove that $\sqrt{||a+b||^2+||c+d||^2} \leq \sqrt{||a||^2+||b||^2} + \sqrt{||c||^2+||d||^2}$, when $a,b,c,d$ are vectors in a Hilbert space $H$.
My attempt:
$$ ||a+b||^2=||a||^2+||b||^2+(||a+b||^2-||a-b||^2)/2 $$ $$ ||c+d||^2=||c||^2+||d||^2+(||c+d||^2-||c-d||^2)/2 $$
where I have used the fact that $<a,b>=\frac{||a+b||^2-||a-b||^2}{4}$
Then, I can write the LHS to be,
$$ \sqrt{ ||a||^2+||b||^2+(||a+b||^2-||a-b||^2)/2 + ||c||^2+||d||^2+(||c+d||^2-||c-d||^2)/2 } $$ I think I need to use $\sqrt{x+y} \leq \sqrt{x} + \sqrt{y}$ when $x,y \geq 0$, but I cannot proceed from the above step. Can someone help?