<s>[INST] Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$ Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality : 
$$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$
Where : 
$s : \operatorname{Semiperimeter}$
$R : \operatorname{circumradius}$ 
I know the relation : 
$$m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$$
But when I multiple together I dont get simple formulas!
So, I need help finding a solution.
Thanks!
 [/INST] In the standard notation we need to prove that:
$$\frac{1}{8}\sqrt{\prod_{cyc}(2a^2+2b^2-c^2)}\leq\frac{1}{2}\cdot\frac{abc}{4S}\cdot\frac{(a+b+c)^2}{4}$$ or
$$a^2b^2c^2(a+b+c)^3\geq\prod_{cyc}(2a^2+2b^2-c^2)\prod_{cyc}(a+b-c).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $$\prod_{cyc}(2a^2+2b^2-c^2)=\prod_{cyc}(2(a^2+b^2+c^2)-3c^2)=$$
$$=8(9u^2-6v^2)^3-12(9u^2-6v^2)^3+18(9u^2-6v^2)(9v^4-6uw^3)-27w^6=$$
$$=27(-w^6+2(3u^2-2v^2)(9v^4-6uw^3)-4(3u^2-2v^2)^3).$$
Also, $$\prod_{cyc}(a+b-c)=\prod_{cyc}(3u-2c)=27u^3-54u^3+36uv^2-8w^3=$$
$$=-8w^3-27u^3+36uv^2.$$
Thus, we need to prove that $f(w^3)\geq0,$ where 
$$f(w^3)=u^3w^6-(-w^6+2(3u^2-2v^2)(9v^4-6uw^3)-4(3u^2-2v^2)^3)(-8w^3-27u^3+36uv^2).$$
But $$f''(w^3)=2u^3-2(-2w^3+2(3u^2-2v^2)(-6u))(-8)+$$
$$-(-8w^3-27u^3+36uv^2)(-2)=-4(157u^3-114uv^2+12w^3)<0,$$ which says that $f$ is a concave function.
Thus, it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.


*

*$w^3\rightarrow0^+$.
in this case the inequality is obvious;

*$\prod\limits_{cyc}(a+b-c)\rightarrow0^+$.
It's obvious again;


*Two variables are equal.


Since our inequality is symmetric and homogeneous, it's enough to assume $b=c=1$.
Thus, $0<a<2$ and we need to prove that
$$a^2(a+2)^3\geq(2a^2+1)^2(4-a^2)a^2(2-a)$$ or
$$(a+2)^2\geq(2a^2+1)^2(2-a)^2$$ or
$$a+2\geq(2a^2+1)(2-a)$$ or
$$a(a-1)^2\geq0$$ and we are done!
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