How do we prove that $a+b+\frac12 \geq \sqrt{a} + \sqrt{b}$ with $a\geq 0$ and $b\geq 0$?

Question

How do we prove that $$a+b+\frac12 \geq \sqrt{a} + \sqrt{b}$$ with $a\geq 0$ and $b\geq 0$?

I tried using Cauchy inequality but doesn't seem to work. Factor things out also seem impossible or require complex numbers.

Answer 1

Let $a,b\geq 0$, then $\sqrt{a},\sqrt{b}\in\mathbb{R}$ and therefore, $$\left(\sqrt{a}-\frac12\right)^2+\left(\sqrt{b}-\frac12\right)^2\geq 0$$ $$\implies a+b-(\sqrt{a}+\sqrt{b})+\frac12\geq 0$$ $$\implies a+b+\frac12\geq \sqrt{a}+\sqrt{b}$$