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Inegalitate conditionata, G.M. 2/2009

Posted: Thu May 21, 2009 5:16 pm
by Claudiu Mindrila
Se considera numerele reale strict pozitive \( x,\ y,\ z \) cu \( x^{4}+y^{4}+z^{4}=3 \). Sa se arate ca: \( \frac{x+y}{\left(x^{2}\sqrt{y}+y^{2}\sqrt{x}\right)^{2}}+\frac{y+z}{\left(y^{2}\sqrt{z}+z^{2}\sqrt{x}\right)^{2}}+\frac{z+x}{\left(z^{2}\sqrt{x}+x^{2}\sqrt{z}\right)^{2}}\ge\frac{3}{2} \).

Catalin Cristea, G.M. 2/2009

Posted: Thu May 21, 2009 5:43 pm
by BogdanCNFB
\( \sum\frac{x+y}{(x^2\sqrt{y}+y^2\sqrt{x})^2}\ge\sum\frac{x+y}{(x^4+y^4)(x+y)}=\sum\frac{1}{x^4+y^4}\ge\frac{9}{2 \sum x^4}=\frac{9}{6}=\frac{3}{2} \).