Inegalitate conditionata cu o inegalitate

[Marius Mainea]

Fie x,y,z >0 astfel incat \( x+y+z\ge x^3+y^3+z^3 \). Sa se arate ca :

\( \frac{x}{x^2+1}(\frac{z}{y})^2+\frac{y}{y^2+1}(\frac{x}{z})^2+\frac{z}{z^2+1}(\frac{y}{x})^2\ge \frac{x+y+z}{2}. \)

[Marius Mainea]

Indicatie: \( \rightarrow \) CBS

[Marius Mainea]

\( \frac{x}{x^2+1}(\frac{z}{y})^2=\frac{(\frac{zx}{y})^2}{x^3+x} \)

[Claudiu Mindrila]

Solutie.
Avem: \( LHS=\sum\frac{\left(\frac{xy}{z}\right)^{2}}{x^{3}+x}\geq\frac{\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^{2}}{x^{3}+y^{3}+z^{3}+x+y+z+}\geq\frac{\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^{2}}{2\left(x+y+z\right)}\geq\frac{3\left(x^{2}+y^{2}+z^{2}\right)}{2\left(x+y+z\right)}\geq\frac{\left(x+y+z\right)^{2}}{2\left(x+y+z\right)}=\frac{x+y+z}{2}=RHS \).

Observatie. Am folosit inegalitatea Cauchy-Schwarz si faptul ca pentru orice \( a,b,c\in\mathbb{R} \) are loc inegalitatea: \( 3\left(a^{2}+b^{2}+c^{2}\right)\geq\left(a+b+c\right)^{2}\geq3\left(ab+bc+ca\right) \)