T
thopeo_kool


Cho a;b;c > 0. Chứng minh rằng :
$\dfrac{5b^3 - a^3}{ab + 3b^2} + \dfrac{5c^3 - b^3}{bc + 3c^2} + \dfrac{5a^3 - c^3}{ca + 3a^2} \le a + b + c$
$\dfrac{5b^3 - a^3}{ab + 3b^2} + \dfrac{5c^3 - b^3}{bc + 3c^2} + \dfrac{5a^3 - c^3}{ca + 3a^2} \le a + b + c$