Inegalitate cu x, y, z

[alex2008]

Fie \( x,y,z\in\mathbb{R}_+ \) . Aratati ca :

\( (x^2+\frac{3}{4})(y^2+\frac{3}{4})(z^2+\frac{3}{4})\ge \sqrt{(x+y)(y+z)(z+x)} \)


Mathematica - Modus Vivendi , 2006

[Marius Mainea]

\( LHS=\prod{(x^2+\frac{3}{4})}=\prod{(x^2+\frac{1}{4})+\frac{1}{2}}\ge\prod{\frac{1}{2}(x+\frac{1}{2})^2+\frac{1}{2}}=\frac{1}{8}\sqrt{\prod{[(x+\frac{1}{2})^2+1][1+(y+\frac{1}{2})^2]}}\ge\frac{1}{8}\sqrt{\prod{(x+y+1)^2}}=\prod{\frac{x+y+1}{2}}\ge\prod{\sqrt{(x+y)}}=RHS \)