Inegalitate

[BogdanCNFB]

Sa se arate ca pentru orice numere reale \( a,b,c>0 \) avem:
\( a+b+c+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge 4\sqrt[3]{abc} \).

[Marius Mainea]

LHS\( \ge a+b+c+\frac{9abc}{(a+b+c)^2}\ge4\sqrt[4]{\frac{(a+b+c)abc}{3}}\ge 4\sqrt[3]{abc} \)