geometry inequality

Question

$M$ is a point in $\triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.

Proof that: $$AA_1 \times BB_1 \times CC_1 \geq 27 (MA_1 \times MB_1 \times MC_1)$$

Answer 1

Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:

$27abc\leq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.

Answer 2

$$ \Leftrightarrow S_{AMB}\cdot S_{BMC}\cdot S_{CMA}\le \left(\dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} \right )^3 = \dfrac{1}{27}S^3 $$

Answer 3

By AM-GM $$\frac{MA_1\cdot MB_1\cdot MC_1}{AA_1\cdot BB_1\cdot CC_1}=\frac{S_{\Delta BMC}}{S_{\Delta ABC}}\cdot\frac{S_{\Delta AMC}}{S_{\Delta ABC}}\cdot\frac{S_{\Delta AMB}}{S_{\Delta ABC}}\leq$$ $$\leq\left(\frac{\frac{S_{\Delta BMC}}{S_{\Delta ABC}}+\frac{S_{\Delta AMC}}{S_{\Delta ABC}}+\frac{S_{\Delta AMB}}{S_{\Delta ABC}}}{3}\right)^3=\frac{1}{27}.$$