Inegalitate conditionata 3

[Marius Mainea]

Fie x,y,z>0 astfel incat xy+yz+zx=1. Demonstrati ca

\( \frac{1}{x+x^3}+\frac{1}{y+y^3}+\frac{1}{z+z^3}\ge \frac{9\sqrt{3}}{4} \)

[maxim bogdan]

Avem: \( S=\sum_{cyc}\frac{1}{x+x^3}=\sum_{cyc}\frac{1}{x(1+x^2)}=\sum_{cyc}\frac{1}{x}\cdot\frac{1}{(x+y)(x+z)} \)

WLOG \( x\geq y\geq z\Rightarrow \frac{1}{x}\leq\frac{1}{y}\leq\frac{1}{z} \) si \( \frac{1}{1+x^2}\leq\frac{1}{1+y^2}\leq\frac{1}{1+z^2}. \)

Acum, din Inegalitatea lui Cebasev ne rezulta ca:

\( 3S\geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{(x+y)(x+z)}+\frac{1}{(y+z)(y+x)}+\frac{1}{(z+x)(z+y)})=\frac{1}{xyz}\cdot\frac{2(x+y+z)}{(x+y)(y+z)(z+x)}\geq\frac{27\sqrt{3}}{4} \)

\( \Leftrightarrow 8(x+y+z)\geq 27\sqrt{3}xyz(x+y)(y+z)(z+x)=\sqrt{3}\cdot 27(1-xy)(1-yz)(1-zx). \)

Din Inegalitatea mediilor obtinem ca : \( 27(1-xy)(1-yz)(1-zx)\leq(1-xy+1-yz+1-zx)^3=8 \)

Mai ramane sa demonstram ca: \( x+y+z\geq \sqrt{3}\Leftrightarrow (x+y+z)^2\geq 3=3(xy+yz+zx) \) ceea ce este evident. Deci inegalitatea noastra e demonstrata!

[Marius Mainea]

\( LHS=\sum{\frac{1}{x(1+x^2)}}=\frac{1}{xyz}\cdot\sum{\frac{yz}{1+x^2}}\ge \frac{1}{xyz}\cdot\frac{(xy+yz+zx)^2}{xy+yz+zx+xyz(xy+yz+zx)}\ge \frac{1}{xyz}\frac{1}{1+\frac{(xy+yz+zx)^2}{3}}\ge \frac{1}{\frac{1}{3\sqrt{3}}}\frac{1}{1+\frac{1}{3}}=RHS \)