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?

By Minkowski (triangle inequality) we obtain:
$$\sum_{cyc}\sqrt{2a^2+b+1}=\sum_{cyc}\sqrt{2a^2+\frac{b(a+b)}{2}+\frac{(a+b)^2}{4}}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{9a^2+4ab+3b^2}=\frac{1}{2}\sum_{cyc}\sqrt{7a^2+b^2+8}\geq$$
$$\geq\frac{1}{2}\sqrt{7(a+b)^2+(b+a)^2+8(1+1)^2}=4.$$
We see that our inequality is true for all reals $a$ and $b$ such that $a+b=2$.
