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Sa se demonstreze ca pentru orice\( x,y,z\in[0,1) \) are loc inegalitatea :

\( \frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}\ge (1+\frac{x+y+z}{3})(\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}). \)

Indicatie:

\( xy\le \frac{x^2+y^2}{2} \)

Din Inegalitatea mediilor obtinem ca:

\( \sum_{cyc}\frac{1}{1-xy}\leq\sum_{cyc}\frac{1}{1-\frac{x^2+y^2}{2}}=\sum_{cyc}\frac{2}{(1-x^2)+(1-y^2)}=\sum_{cyc}\frac{2}{\frac{1}{\frac{1}{1-x^2}}+\frac{1}{\frac{1}{1-y^2}}}\leq\sum_{cyc}\frac{\frac{1}{1-x^2}+\frac{1}{1-y^2}}{2}=\sum_{cyc}\frac{1}{1-x^2}. \) Am aplicat Inegalitatea mediilor (AM-HM).

Mai ramane sa demonstram ca:

\( (\sum_{cyc}\frac{1}{1-x^2})(\sum_{cyc}1+x)\leq 3(\sum_{cyc}\frac{1}{1-x}) \)

WLOG: \( x\geq y\geq z \). Deci tripletele: \( (\frac{1}{1-x^2};\frac{1}{1-y^2};\frac{1}{1-z^2}) \) si \( (1+x;1+y;1+z) \) sunt la fel ordonate.

Deci din Inegalitatea lui Cebasev:

\( \Longrightarrow (\sum_{cyc}\frac{1}{1-x^2})(\sum_{cyc}1+x)\leq 3(\sum_{cyc}\frac{1+x}{1-x^2})=3(\sum_{cyc}\frac{1}{1-x}). \)

Cred ca era \( \sum\frac{2}{\left(1-x^{2}\right)+\left(1-y^{2}\right)}\leq\sum\frac{2}{\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}}} \) .

E corecta solutia! Am folosit inegalitatea mediilor sub forma:

\( \frac{2}{\frac{1}{a}+\frac{1}{b}}\leq\frac{a+b}{2} \), unde \( a=\frac{1}{1-x^2} \) si \( b=\frac{1}{1-y^2}. \)