# Prove that $\log_{2b+c}a+\log_{2c+a}b+\log_{2a+b}c\ge\frac{3}{2}$

Let $a\ge 3,b\ge3,c\ge3$. Prove that: $$\log_{2b+c}a+\log_{2c+a}b+\log_{2a+b}c\ge\frac{3}{2}$$

I don't know what to do. Rewrite to $\ln(x)$ or $e^x$ But it's not work

-
The inequality is an equality for $a=b=c=3$. So maybe showing that the left hand side is an increasing function of $a$, $b$, and $c$ (if such a result is indeed true $\ldots$ ) – Dilip Sarwate Sep 19 '12 at 1:43
My solution: $\frac{2}{c}+\frac{1}b\le1$ so $2b+c\le bc$. Hence $log_{2b+c}a=\frac{lna}{ln(2b+c)}\ge\frac{lna}{ln(bc)}=\frac{lna}{lnb+lnc}$. then we use Nesbit inequality – tangkhaihanh Sep 19 '12 at 14:39

My solution.

Consider the right side of the expression the function $f(a,b,c); a\ge 3,b\ge 3,c\ge 3$

So you have

$$f(a,b,c)=\log_{2b+c}{a}+\log_{2c+a}{b}+\log_{2a+b}{c}$$

So you are trying to prove that

$$f(a,b,c)\ge \frac{3}{2}$$

Consider the instance of the minimum value for all variables (3).

$$f(3,3,3)=\log_9{3}+\log_9{3}+\log_9{3}$$ $$=3\log_9{3}$$ $$=3\left(\frac{1}{2}\right)$$ $$=\frac{3}{2}$$

So you have shown that for the minimum values of all variables, the function is equal to the minimum value of the right hand side. Now you have to establish that for each variable $[a,b,c]$ increasing, the function as a whole is increasing.

I would recommend taking the first or second partial derivative to show the slope is increasing for each variable...

-