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For $a,b,c>0$ and $abc=8$. Find max : $\frac{1}{2a+b+6}+\frac{1}{2b+c+6}+\frac{1}{2c+a+6}$

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What have you tried? Looks like the Lagrange multipliers would do the job here. – fuglede Jan 4 '14 at 14:56
up vote 3 down vote accepted

Let $a=2x^2,b=2y^2,c=2z^2$.Without loss of generality: $x,y,z>0,xyz=1$.

From the AM-GM Inequality, $\frac 1{2a+b+6}=\frac 1{2(2x^2+y^2+3)}=\frac 1{2(x^2+1+x^2+y^2+2)}\le \frac 1{4(x+xy+1)}$,

Similarly, $\frac 1{2b+c+6}\le \frac 1{4(y+yz+1)}=\frac x{4(xy+1+x)}$,

$\frac 1{2c+a+6}\le \frac 1{4(z+zx+1)}=\frac {xy}{4(1+x+xy)}$,

We have $P\le \frac 14$. The equality occurs when $a=b=c=2$

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