Medii

[Claudiu Mindrila]

Sa se arate ca daca \( a,b\in (0, \infty) \), atunci
\( \frac{8a^2b^2}{(a^2+1)(b^2+1)}\leq \sqrt{(a^4+1)(b^4+1)} \).
Nistor Budescu, R.M.T. 4/2008

[Laurian Filip]

\( \frac{8a^2b^2}{(a^2+1)(b^2+1)}\leq \frac{8a^2b^2}{4ab}=2ab=\sqrt{4a^2b^2}\leq \sqrt{(a^4+1)(b^4+1)} \)