A problem :$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$ Hello during a problem I have this to solve :
Let $a,b,c,d,e,f$ be real positiv number such that $a\geq b\geq c\geq d\geq e\geq f\geq 1$ then :
$$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$$
I did not find any counter-example (with Pari-Gp and Wolfram alpha) so maybe it's true 
Advice :I do not think that Am-Gm will works on it .
Thanks a lot

This inequality simply follows from the original inequality for $n=3$, so I guess its proof won't be easier. Indeed, dividing both sides by $def$ we transform it to 
$$(def)^2(a+b+c+d+e+f)^3\geq abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\right)^3.$$    
Now we note that with $a$, $b$, and $c$ fixed the left hand side is at least $(a+b+c+3)^3$, the right hand side is at most $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3\right)^3$ and both sides are equal to their extremal values when $d=e=f=1$.
