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?

$2a^2+b+2=2a^2+3-a=(\frac{a^2+1}{2}-a)+(\frac{3a^2+5}{2})\geq\frac{3a^2+5}{2}$.
On the other hand by QM-AM $\sqrt{\frac{a^2+a^2+a^2+1+1+1+1+1}{8}}\geq\frac{3a+5}{8}$, therefore $\sqrt{3a^2+5}\geq\frac{3a+5}{\sqrt{8}}$. Now we return back: $\sqrt{2a^2+b+2}\geq\sqrt{\frac{3a^2+5}{2}}\geq\frac{3a+5}{\sqrt{2\times8}}=\frac{3a+5}{4}$. 
Therefore $\sqrt{2a^2+b+2}+\sqrt{2b^2+a+2}\geq\frac{(3a+5)+(3b+5)}{4}=\frac{3(a+b)+10}{4}=\frac{16}{4}=4$
