<s>[INST] Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$

My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$
Or $\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$
Then i can't prove $\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$
I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QM\ge AM+GM$ with $n=2$ so is that true for $n=i$?
 [/INST] An alternative approach, let
$$
u = \frac{a + b}{2} \quad \text{and} \quad v = ab
$$
so that the inequality becomes
$$
\frac{v}{u} + \sqrt{2u^2 - v} \ge \sqrt{v} + u.
$$
Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to
$$
x \ge \sqrt{1 + x} - \sqrt{1 - x} \quad (\text{for}\,\, x \le 0)
$$
which is shown fairly easily. The condition comes from
$$
u = 1 \implies b = 2 - a \implies x = a(2 - a) - 1 = -(a-1)^2 \le 0.
$$
</s>