O interesanta inegalitate cu mediane

[Virgil Nicula]

$\boxed{\mathrm{Sa\ se\ demonstreze\ ca\ in\ orice}\ \triangle ABC\ \mathrm{exista\ inegalitatea}\ 4m_bm_c\le 2a^2+bc}$ (notatii standard).

[mihai miculita]

$4m_bm_c\le2a^2+bc\Leftrightarrow 4\sqrt{\dfrac{2.(a^2+c^2)-b^2}{4}.\dfrac{2.(a^2+b^2)-c^2}{4}}\le2a^2+bc\Leftrightarrow \sqrt{[2.(a^2+c^2)-b^2].[2.(a^2+b^2)-c^2]}\le2a^2+bc |()^2\Leftrightarrow$
$\Leftrightarrow [2.(a^2+c^2)-b^2].[2.(a^2+b^2)-c^2]\le(2a^2+bc)^2\Leftrightarrow4(a^2+c^2)(a^2+b^2)-2c^2.(a^2+c^2)-2b^2.(a^2+b^2)+b^2c^2\le4a^4+2a^2bc+b^2c^2\Leftrightarrow$
$\Leftrightarrow 4a^4+4a^2b^2+4a^2c^2+4b^2c^2-2a^2c^2-2c^4-2a^2b^2-2b^4\le4a^4+2a^2bc+b^2c^2\Leftrightarrow 2a^2b^2+2a^2c^2+4b^2c^2\le2b^4+4a^2bc+2c^4 |:2\Leftrightarrow$
$\Leftrightarrow a^2b^2+a^2c^2+2b^2c^2\le b^4+2a^2bc+c^4\Leftrightarrow 0\le(b^4-2b^2c^2+c^4)-a^2(b^2-2bc+c^2)\Leftrightarrow0\le(b^2-c^2)^2-a^2(b-c)^2\Leftrightarrow$
$\Leftrightarrow 0\le (b-c)^2(b+c)^2-a^2(b-c)^2\Leftrightarrow 0\le (b-c)^2[(b+c)^2-a^2]\Leftrightarrow\boxed{ 0\le (b-c)^2(b+c-a)(b+c+a)}.$ Adevarata in orice triunghi $ABC$.
Evident, in inegaltatea data, egalitatea are loc doar in cazul $\boxed{b=c}$.