# Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$

If $A=\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}$ and $B = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ and $a,b,c>0.$ Then prove that $A\geq 1+B$

$\bf{My\; Try::}$We can write $A$ and $B$ as $$=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} = \frac{3}{A}$$ and $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{3}{B}$$

Using $\bf{cauchy \; schwarz }$ Inequality

$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geq \frac{3^2}{1+a+1+b+1+c} = \frac{9}{3+a+b+c}$$

Now How can i solve after that , Help Required, Thanks

• It can be shown that any power mean with exponent $< 1$ is super-additive, i.e. $$M_p(a+1,b+1,c+1)\geq 1+M_p(a,b,c)$$ and in this case we are just dealing with the $p=-1$ case (harmonic mean). – Jack D'Aurizio Sep 24 '16 at 12:03
• You can get bold letters without math mode by using asterisks: **bold letters** – Arthur Sep 25 '16 at 11:51
• Thanks Jack D'Aurizio , Would you like to explain me the meaning of first line – juantheron Sep 26 '16 at 6:52

we need to prove that $$\frac{3(1+a)(1+b)(1+c)}{\sum\limits_{cyc}(ab+2a+1)}\geq1+\frac{3abc}{ab+ac+bc}$$ or $$\sum\limits_{cyc}(2a^2b^2-2a^2bc+a^2b+a^2c-2abc)\geq0$$ or $$\sum\limits_{cyc}(a-b)^2(c^2+c)\geq0$$ Done!
We'll rewrite our inequality in the following form $$3\sum\limits_{cyc}\frac{1}{a(1+a)}\geq\sum\limits_{cyc}\frac{1}{1+a}\sum\limits_{cyc}\frac{1}{a}$$ which is Rearrangement.
• $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ and $\left(\frac{1}{1+a},\frac{1}{1+b},\frac{1}{1+c}\right)$ are the same ordered. Thus, $\sum\limits_{cyc}\frac{1}{a(1+a)}$ is a biggest sum. – Michael Rozenberg Sep 24 '16 at 8:15