Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$ 
Show that for any two positive real numbers $a$ and $b$,  $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$

My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^2\geq2ab \\\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}$
I do not know how to link them together. Appreciate any tips. 

By setting:
$$ M_p(a,b)=\lim_{q\to p^+}\left(\frac{a^q+b^q}{2}\right)^{\frac{1}{q}}$$
we have to prove: $$2M_1\geq M_0+M_2, $$ 
but $M_0^2+M_2^2 = 2M_1^2$, hence the previous line follows from Cauchy-Schwarz inequality:
$$ M_0+M_2 \leq \sqrt{2}\sqrt{M_0^2+M_2^2} = 2M_1.$$
