Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$

Question

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!

Answer 1

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.

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

2. $\prod\limits_{cyc}(a+b-c)\rightarrow0^+$.

It's obvious again;

3. 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!

Answer 2

Note that the triangles ABD and EDC are similar. Then,

$$\frac{AD}{BD}=\frac{CD}{ED}\implies \frac{m_a}{\frac a2}=\frac{\frac a2}{AE-m_a} \implies m_a^2 -AE\cdot m_a + \frac {a^2}4=0$$

which, since $AE \le 2R$ and $a=2R\sin A$, leads to

$$m_a =\frac12(AE+\sqrt{AE^2-a^2})\le \frac12\left[2R+\sqrt{(2R)^2-(2R\sin A)^2}\right] =2R\cos^2 \frac A2$$ Likewise, $m_b\le 2R\cos^2 \frac B2$ and $m_c\le 2R\cos^2 \frac C2$. Together, we have

$$\begin{align} m_a m_bm_c & \le \frac12R^3\left( 4\cos\frac A2\cos \frac B2\cos \frac C2\right)^2 \\ & = \frac12R^3\left( 2 \cos\frac A2 \left(\cos \frac {B+C}2+\cos \frac {B-C}2 \right)\right)^2 \\ & = \frac12R^3\left( 2 \cos\frac A2 \sin\frac A2+2 \sin\frac {B+C}2\cos \frac {B-C}2 \right)^2 \\ & = \frac12R^3\left( \sin A + \sin B + \sin C \right)^2 \\ & = \frac12R^3 \left( \frac a{2R} + \frac b{2R} + \frac c{2R}\right)^2 \\ & = \frac12R \left( \frac{a + b+ c}2 \right)^2 \\ \end{align}$$

Thus,

$$m_a m_bm_c \le \frac12Rs^2$$