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 $n=i$?
 A: This Problem is well known.
Note that $$\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)^2-4ab}{2(a+b)}$$
after squaring two times you will get
$$\left(\frac{(a+b)^2}{2}-\frac{(a-b)^4}{4(a+b)^2}\right)^2\geq 4ab\left(\frac{a^2+b^2}{2}\right)$$
Doing this Algebra you will get
$$\frac{1}{16}\frac{(a-b)^8}{(a+b)^4}\geq 0$$ which is true.
A: Note that $$\begin{align}\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}&=\frac{\left(\frac{a^2+b^2}{2}\right)-ab}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}=\frac{2\left(\frac{a-b}{2}\right)^2}{\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}}
\end{align}$$
If we can show that
$$\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b\,,$$
then
$$\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\geq \frac{2\left(\frac{a-b}{2}\right)^2}{a+b}=\frac{a+b}{2}-\frac{2ab}{a+b}\,,$$
which is equivalent to what we need to prove.
Note that $$\begin{align}
\sqrt{\frac{a^2+b^2}{2}}&=\sqrt{\left(\frac{a+b}{2}\right)^2+\left(\frac{a-b}{2}\right)^2}
\\
&=\sqrt{\left(\frac{a+b}{2}\right)^2+\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^2\,\left(\frac{\sqrt{a}+\sqrt{b}}{\sqrt2}\right)^2}
\\
&=\sqrt{\left(\frac{a+b}{2}\right)^2+\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^2\left(\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^2+2\sqrt{ab}\right)}
\\
&=\sqrt{\left(\frac{a+b}{2}\right)^2+2\sqrt{ab}\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^2+\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^4}
\\
&\leq\sqrt{\left(\frac{a+b}{2}\right)^2+2\left(\frac{a+b}{2}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^2+\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^4} 
\\
&=\frac{a+b}{2}+\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt2}\right)^2\,.\end{align}$$
Thus,
$$\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b\,.$$
A: 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
$$
u = 1 \implies b = 2 - a \implies x = a(2 - a) - 1 = -(a-1)^2 \le 0.
$$
A: I think this is the simplest solution:
So you are stuck here:
$$\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$$
...which is equivalent to:
$$\sqrt\frac{a^2+b^2}{2}+\sqrt{ab}\le a+b\tag{1}$$
Apply Jensen inequality:
$$\frac{f(x_1)+...+f(x_n)}{n}\le f(\frac{x_1+...+x_n}{n})$$
...for concave function $f(x)=\sqrt{x}$ and $x_1=\frac{a^2+b^2}{2}$, $x_2=ab$:
$$\frac{\sqrt\frac{a^2+b^2}{2}+\sqrt{ab}}2\le \sqrt{\frac{\frac{a^2+b^2}{2}+ab}{2}}\tag{2}$$
Simplify (2) and you get (1) directly.
A: Thanks everyone i had the answer for my stuck.  
We have : $$LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$$
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 need to prove  $$\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$$
Or $\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b$. Which's true by C-S $$\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)^2\le 2\left(\frac{a^2+b^2}{2}+ab\right)=\left(a+b\right)^2$$
A: Let $a^2+b^2=2t^2ab$, where $t>0$.
Thus, by AM-GM $t\geq1$ and we need to prove that
$$\frac{2ab}{\sqrt{a^2+b^2+2ab}}+\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}+\frac{\sqrt{a^2+b^2+2ab}}{2}$$ or
$$\sqrt{\frac{2}{t^2+1}}+t\geq1+\sqrt{\frac{t^2+1}{2}}$$ or
$$t-1\geq\sqrt{\frac{t^2+1}{2}}-\sqrt{\frac{2}{t^2+1}}$$ or
$$t-1\geq\frac{t^2-1}{\sqrt{2(t^2+1)}}$$ or
$$\sqrt{2(t^2+1)}\geq t+1,$$ which is true by C-S:
$$\sqrt{2(t^2+1)}=\sqrt{(1^2+1^2)(t^2+1)}\geq t+1.$$
Done!
