How to prove $\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$

Question

If $a,b,c>0$ prove that $$\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$$

It is stronger than the well-known result $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.$$

I tried to verify the right side as $$3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2=\frac{1}{3}\left(9+\sum_{cyc}\frac{(b+c)^2-4bc}{(b+c)^2}\right)$$ $$=\frac{4}{3}\left(3-\sum_{cyc}\frac{bc}{(b+c)^2}\right)=\frac{4}{3}\sum_{cyc}\frac{b^2+c^2+bc}{(b+c)^2}$$

From here, I don't know how to continue. Can you help me?

Answer 1

Proof.

By using Cauchy-Schwarz and AM-GM inequality \begin{align*} \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}&=\sqrt{\frac{(a+b+c)(a^2(b+c)+abc)}{a^2bc}}\\&\ge \frac{a(b+c)+a\sqrt{bc}}{a\sqrt{bc}}\\&=2\cdot\frac{b+c}{2\sqrt{bc}}+\frac{4bc}{(b+c)^2}+1-\frac{4bc}{(b+c)^2}\\&\ge 3\sqrt[3]{\frac{b+c}{2\sqrt{bc}}\cdot \frac{b+c}{2\sqrt{bc}}\cdot\frac{4bc}{(b+c)^2}}+1-\frac{4bc}{(b+c)^2}\\&=4-\frac{4bc}{(b+c)^2}=\frac{4(b^2+c^2+bc)}{(b+c)^2}. \end{align*} Now, by your try we end the proof. Equality holds at $a=b=c>0.$

Answer 2

Here is my solution.

Step 1

Substitute $x=\frac{b+c}{\sqrt{bc}}$, $x=\frac{c+a}{\sqrt{ca}}$, $z=\frac{a+b}{\sqrt{ab}}$

Then the problem changes as below.

$$ \sqrt{\sum_{cyc} x^2 - 3} \geq 4 - \frac{4}{3} \sum_{cyc} \frac{1}{x^2}$$

Step 2

To get rid of $\sum_{cyc} \frac{1}{x^2}$, note that $t+\frac{4}{t^2} \geq 3$ for every $t>0$. This implies the following.

$$ 1+\frac{1}{3}\sum_{cyc} x \geq 4 - \frac{4}{3} \sum_{cyc} \frac{1}{x^2}$$

Therefore, it suffices to prove the following.

$$ \sqrt{\sum_{cyc} x^2 - 3} \geq 1+\frac{1}{3}\sum_{cyc} x $$

Step 3

Note that C-S and AM-GM inequalities imply $\sum_{cyc} x^2 \geq \frac{1}{3} (\sum_{cyc} x)^2$ and $\sum_{cyc} x \geq 6$. Combining these two and a littl bit of algebra, we get the following.

$$ \sqrt{\sum_{cyc} x^2 - 3} \geq \sqrt{\frac{1}{3}(\sum_{cyc} x)^2 - 3} \geq 1+\frac{1}{3}\sum_{cyc} x $$

It is proved. The equality holds if and only if $x=y=z=2$ or $a=b=c$.

Answer 3

By the d8g3n1v9's beautiful idea we can prove a stronger inequality:

let $a$, $b$ and $c$ be positive numbers. Prove that: $$\sqrt{(a+b+c)\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}\right)}\geq3+\tfrac{2}{3}\left(\left(\tfrac{a-b}{a+b}\right)^2+\left(\tfrac{b-c}{b+c}\right)^2+ \left(\tfrac{c-a}{c+a}\right)^2\right).$$

Indeed, let $\frac{a+b}{\sqrt{ab}}=2\sqrt{z},$ $\frac{b+c}{\sqrt{bc}}=2\sqrt{x}$ and $\frac{c+a}{\sqrt{ca}}=2\sqrt{y}.$

Thus, $x\geq1$, $y\geq1$ and $z\geq1$ and we need to prove that: $$\sqrt{4(x+y+z)-3}\geq3+\frac{2}{3}\sum_{cyc}\frac{x-1}{x}$$ or $$\sqrt{4(x+y+z)-3}+\frac{2}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq5.$$

Now, let $x+y+z=3u$.

Thus, $u\geq1$ and by C-S $$\sqrt{4(x+y+z)-3}+\frac{2}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq\sqrt{4(x+y+z)-3}+\frac{6}{x+y+z}=$$ $$=\sqrt{12u-3}+\frac{2}{u}\geq5,$$ where the last inequality it's $$(u-1)^2(3u-1)\geq0$$ and we are done.