\frac{a^{3}+b^{3}}{2}\geq \frac{(a^{2}+b^{2})^{3}}{(a+b)^{2}}
Edited by minh04042006, 22-01-2018 - 16:22.
\frac{a^{3}+b^{3}}{2}\geq \frac{(a^{2}+b^{2})^{3}}{(a+b)^{2}}
Edited by minh04042006, 22-01-2018 - 16:22.
0 members, 1 guests, 0 anonymous users