Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?

Suppose without loss of generality that $a\geqslant b$ so that $0\leqslant b \leqslant 1$ and write $a = 2- b$.
Our expression becomes
$$\sqrt{2(2-b)^2 + b + 1} + \sqrt{2b^2 - b + 3}$$
and this can be minimized for $b\in[0,1]$ with usual calculus techniques.
