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Sa se arate ca oricare ar fi \( a,b,c\in(0,\infty) \) are loc inegalitatea

\( \frac{ab}{a+b+1}+\frac{bc}{b+c+1}+\frac{ca}{c+a+1}\le \frac{\sqrt{3}}{8}\sqrt{(a+b)^3+(b+c)^3+(c+a)^3}. \)

Cristian Grecu , RMT/4/2008

Deoarece \( a+b+1=a+\frac{1}{2}+b+\frac{1}{2}\geq 4\sqrt[4]{\frac{ab}{4}} \) \( \Longrightarrow \frac{ab}{a+b+1}\leq \frac{ab}{4\sqrt[4]{\frac{ab}{4}}}=\frac{\sqrt[4]{a^3b^3}}{2\sqrt{2}} \).
Deci \( LHS=\sum \frac{ab}{a+b+1}\leq \sum \frac{\sqrt[4]{a^3b^3}}{2\sqrt{2}} \).

Apoi \( RHS=\frac{\sqrt{3}}{8}\sqrt{(a+b)^3+(b+c)^3+(c+a)^3} \). Dar deoarece \( a+b\geq 2\sqrt{ab}\Longrightarrow (a+b)^3\geq 8\sqrt{a^3b^3} \). Atunci \( RHS \geq \frac{\sqrt{3}}{8}\sqrt{8}\cdot \sqrt{\sqrt{a^3b^3}+\sqrt{b^3c^3}+\sqrt{c^3a^3}} \).
Sa notam \( x=\sqrt[4]{a^3b^3}, y=\sqrt[4]{b^3c^3}, z=\sqrt[4]{c^3a^3}. \)

Avem: \( LHS \leq RHS \Longleftrightarrow \frac{1}{2\sqrt{2}}(x+y+z) \leq \frac{\sqrt{3(x^2+y^2+z^2)}}{2\sqrt{2}} \), evident.