Question

$\text{Sa se arate ca} ~\forall ~x,y,z\in\mathbb{Q+}~ a.i. ~x\cdot y\cdot z=1, \text{are loc inegalitatea:}$ $\frac{1}{x^3+y^3+1}+ \frac{1}{y^3+z^3+1}+ \frac{1}{z^3+x^3+1} \leq 1.$

Answer 1

$Se~foloseste~inegalizatea~a^3+b^3 \geq ab(a+b)~adevarata~ \forall~a,b \geq 0, \\ \\ Astfel~avem:~ \\ \\ \frac{1}{x^3+y^3+1}= \frac{1}{x^3+y^3+xyz} \leq \frac{1}{xy(x+y)+xyz}= \frac{1}{xy(x+y+z)}= \frac{z}{xyz(x+y+z)}. \\ \\ Si~relatiile~analoage:~ \frac{1}{y^3+z^3+1} \leq \frac{x}{xyz(x+y+z)}~si~ \frac{1}{z^3+x^3+1} \leq \frac{y}{xyz(x+y+z)}.$

$Acum,~insumand~aceste~relatii,~obtinem: \\ \\ \frac{1}{x^3+y^3+1}+ \frac{1}{y^3+z^3+1}+ \frac{1}{z^3+x^3+1} \leq \frac{x+y+z}{xyz(x+y+z)}= \frac{1}{xyz}=1.$