Norm squared inequality

Question

I am reading a paper in optimization and confused about the following identity: Let $a,b,c$ be elements of a real Hilbert space $H$,

$$ \|a-b\|^2 \leq 2\|a-c\|^2 + 2\|c-b\|^2. $$

At first I thought it was just the triangle inequality and adding/subtracting $c$ inside the the first norm. However, since we have the norm squared instead of norm I am not sure that's the case. Can the inequality be true as stated?

Answer 1

At first I thought it was just the triangle inequality and adding/subtracting $c$ inside the the first norm.

That, and then from the obtained

$$\lVert a-b\rVert^2 \leqslant \lVert a-c\rVert^2 + 2\lVert a-c\rVert\cdot\lVert c-b\rVert + \lVert c-b\rVert^2$$

we conclude using the inequality

$$2uv \leqslant u^2 + v^2$$

for $u, v \in \mathbb{R}$, following from $(u-v)^2 \geqslant 0$.

Answer 2

We can you also C-S here: $$2||a-c||^2+2||c-b||^2=(1^2+1^2)\left(||a-c||^2+||c-b||^2\right)\geq$$ $$\geq\left(||a-c||+||c-b||\right)^2\geq||a-c+c-b||^2=||a-b||^2.$$