How to prove the inequality $\sqrt{\frac{a^2+b^2}2}+\frac{a+b}2\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}$? I want to prove that for all $a,b>0$: $$\sqrt{\frac{a^2+b^2}2}+\frac{a+b}2\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}.$$
My attempts: 
Failed attempt. We know by QM-AM inequality that $\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2$ so it would be enough to prove $$a+b\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}$$ but this is wrong, try $a=1$ and $b=4$.

Let
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2} \ge \frac{a^2+b^2}{a+b}+\sqrt{ab}$$
$$\implies \sqrt{\frac{a^2+b^2}{2}}\ge \frac{a^2+b^2}{a+b}-\frac{a+b}{2}+\sqrt{ab}$$
Let $a=1, b=t^2$, then
$$\implies \sqrt{\frac{1+t^4}{2}} \ge \frac{1+t^4}{1+t^2}-\frac{1+t^2}{2}+t=\frac{(1-t^2)^2+2t(1+t^2)}{2(1+t^2)}$$ So now we need to prove that
$$F=2(1+t^4)(1+t^2)^2-[(1-t^2)^2+2t(1+t^2)]^2 \ge 0$$
By expanding it we get 
$$F=t^8-4t^7+4t^6+4t^5-10t^4+4t^3+4t^2-4t+1=(t-1)^2(t+1)^6 \ge 0.$$
Hence proved and the equality hold when $t=1$ $(a=b).$
