Is this following inequality true?
If $m_a$ be the length of the median on side $a$ of acute angled triangle $ABC$ then $(m_a)^2 \le 2b^2c^2(1+\cos A)$..
After simplifying I am getting $ \frac{1}{bc} < \frac{4((b+c)^2-a^2)}{2b^2+2c^2-a^2}$ .But what to do next?
Something is wrong in the given inequality. If we measure lengths in meters, $(m_a)^2$ is measured in squared meters, while $2b^2 c^2(1+\cos A)$ is measured in $m^4$.
However, by Stewart's theorem $$ m_a^2 = \frac{2b^2+2c^2-a^2}{4} = \frac{b^2+c^2+2bc\cos(A)}{4}$$ and since $b^2+c^2\geq 2bc$, $$ m_a^2 \geq \frac{2bc(1+\cos A)}{4} $$ for any triangle $ABC$ with side lengths $a,b,c$.
