# 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 $n=i$?

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, $$\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b\,.$$

• I think you can use Cauchy-Schwarz's inequality for prove $\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\leq a+b\,.$ it's easy than yours, btw thank you. Aug 5, 2018 at 11:40
• Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof. Aug 5, 2018 at 11:41
• By C-S: $\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)^2\le 2\left(\frac{a^2+b^2}{2}+ab\right)=\left(a+b\right)^2$ Done :). Aug 5, 2018 at 11:43
• Good. You can write this as an answer, and I will upvote it. One can answer his own question. Aug 5, 2018 at 11:44

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.

• Thanks Dr.Son now i know your solution. Aug 5, 2018 at 11:41

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

$$u = 1 \implies b = 2 - a \implies x = a(2 - a) - 1 = -(a-1)^2 \le 0.$$

• Nice solutions are written here! Aug 5, 2018 at 11:49

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.

• That is a good solution too. Aug 5, 2018 at 11:48
• Upvote won't hurt :) Aug 5, 2018 at 11:48
• Sorry but votes cast by those with less than 15 reputation are recorded .-. . Aug 5, 2018 at 11:50
• @NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points. Aug 5, 2018 at 11:51

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 $$\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)^2\le 2\left(\frac{a^2+b^2}{2}+ab\right)=\left(a+b\right)^2$$

• This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved. Aug 5, 2018 at 11:55

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!