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
The note of Alex says that it remains to prove that $$(a_1+a_2+a_3+3)^3\geq a_1a_2a_3\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+3\right)^3$$ for all $a_i\geq1$.
Indeed, let $a_1=a^3$, $a_2=b^3$ and $a_3=c^3$.
Thus, it's enough to prove that $$a^3+b^3+c^3+3\geq abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+3\right)$$ or $$a^2b^2c^2(a^3+b^3+c^3+3)\geq a^3b^3+a^3c^3+b^3c^3+3a^3b^3c^3$$ or $$a^2b^2c^2\sum_{cyc}(a^3-abc)\geq\sum_{cyc}(a^3b^3-a^2b^2c^2)$$ or $$a^2b^2c^2(a+b+c)\sum_{cyc}(a-b)^2\geq(ab+ac+bc)\sum_{cyc}c^2(a-b)^2$$ or $$\sum_{cyc}(a-b)^2c^2(a^2b^2(a+b+c)-ab-ac-bc)\geq0.$$ Now, since our inequality is symmetric, we can assume $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a-b)^2c^2(a^2b^2(a+b+c)-ab-ac-bc)\geq$$ $$\geq(a-c)^2b^2(a^2c^2(a+b+c)-ab-ac-bc)+(b-c)^2a^2(b^2c^2(a+b+c)-ab-ac-bc)\geq$$ $$\geq(b-c)^2a^2(a^2c^2(a+b+c)-ab-ac-bc)+(b-c)^2a^2(b^2c^2(a+b+c)-ab-ac-bc)=$$ $$=(b-c)^2a^2(c^2(a^2+b^2)(a+b+c)-2(ab+ac+bc))\geq$$ $$\geq(b-c)^2a^2((a+b)(a+b+c)-2(ab+ac+bc))=(b-c)^2a^2(a^2+b^2-ac-bc)\geq0$$ and we are done!
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 $d=e=f=1$.
