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Demonstrati ca daca x,y,z sunt pozitive atunci

\( \frac{x^2+xy+y^2}{y+z}+\frac{y^2+yz+z^2}{z+x}+\frac{z^2+zx+x^2}{x+y}\ge \frac{3}{2}(x+y+z) \)

OLM,2010-Dambovita

Indicatie: Se inmultesete expresia cu 2 si se aplica de 3 ori inegalitatea C.B.S forma Titu Andreescu

\( 2LHS=\sum{\frac{2(x^2+xy+y^2)}{y+z}}=\sum{\frac{(x+y)^2}{y+z}}+\sum{\frac{x^2}{y+z}}+\sum{\frac{y^2}{y+z}}\ge \frac{(x+y+y+z+z+x)^2}{y+z+z+x+x+y}+\frac{(x+y+z)^2}{y+z+z+x+x+y}+\frac{(y+z+x)^2}{y+z+z+x+x+y}= \)
\( =\frac{4(x+y+z)^2}{2(x+y+z)}+\frac{(x+y+z)^2}{2(x+y+z)}+\frac{(x+y+z)^2}{2(x+y+z)}=\frac{6(x+y+z)^2}{2(x+y+z)}=3(x+y+z)=2RHS \)