Given icoscales triangle with sides $a, b, c, a = b$; median performed to the side of triangle, say, to b, denoted as $m_b.$ Note: I need to maximize area of triangle. I need to solve it using inequalities, not Lagrange multipliers etc.
Firstly i tried to solve it using median formula $m_b^2 = \dfrac{2a^2 + 2c^2 - b^2}{4} = \dfrac{a^2}{4} + \dfrac{c^2}{2}$ and Geron's formula of sqare of triangle: \begin{align} S^2 &= p(p - a)(p-b)(p -c), \; p = \dfrac{2a+c}{2} = a + \dfrac{c}{2}, \\ S^2 &= \Big(a^2 - \dfrac{c^2}{4} \Big) \dfrac{c^2}{4} = \Big(\dfrac{a^2}{4} - \dfrac{c^2}{16}\Big)c^2 = \Big(m^2_b - \dfrac{9c^2}{16}\Big) c^2 = \\ &=\Bigg(\sqrt{\Big(m^2_b - \dfrac{9c^2}{16}\Big) c^2 }\Bigg)^2 \overset{AM-GM}{\leq} \Big(\dfrac{m^2_b}{2}+ \dfrac{7c^2}{32}\Big)^2 \end{align} Then since inequality turns to equality iff $m^2_b - \dfrac{9c^2}{16} = c^2$, i got $c = \dfrac{4}{5}m_b$ and $S_{\max} = \dfrac{16}{25}m^2_b$. Well, it seems wrong answer.
How can i solve it right-way?
You could try completing the square:
\begin{align} S^2 &= \left(m^2_b - \dfrac{9c^2}{16}\right)c^2= m^2_b c^2 - \dfrac{9c^4}{16} \\ &= \left(\dfrac23m^2_b\right)^2 - \left(\dfrac23m^2_b\right)^2 + 2\left(\dfrac23m^2_b\right) \left(\dfrac34c^2\right) - \left(\dfrac34c^2\right)^2 \\ &= \dfrac49m^4_b - \left( \dfrac23m^2_b-\dfrac34c^2\right)^2 \\ &\le \dfrac49m^4_b \end{align}
with equality (i.e. a maximum) when $\frac23m^2_b=\frac34c^2$, i.e. $c= \sqrt{\frac89}m_b$, giving $S_{\max}=\frac23m^2_b$.