Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$.

We may set $a=1-t, b=1+t$ and study the behaviour of $$ f(t)=\sqrt{2(1-t)^2+(1+t)+1}+\sqrt{2(1+t)^2+(1-t)+1}=\sqrt{4-3t+t^2}+\sqrt{4+3t+t^2} $$ over $[-1,1]$. $f(t)$ is even and convex (as the sum of two convex functions), hence it attains its minimum value at the origin, QED.

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 $a+b=2$.