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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

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    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 $\ldots$ )2012-09-19
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    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 inequality2012-09-19

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