Chủ đề này có 1 trả lời

[Kira Tatsuya]

$x,y,z>0$ thỏa $xyz=1$.Tìm $min$:

$P=\frac{x^9+y^9}{x^6+x^3y^3+y^6}+\frac{y^9+z^9}{y^6+y^3z^3+z^6}+\frac{z^9+x^9}{z^6+z^3x^3+x^6}$

 

[520]

$x,y,z>0$ thỏa $xyz=1$.Tìm $min$:

$P=\frac{x^9+y^9}{x^6+x^3y^3+y^6}+\frac{y^9+z^9}{y^6+y^3z^3+z^6}+\frac{z^9+x^9}{z^6+z^3x^3+x^6}$

Ta có $\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}=a+b-\frac{2ab(a+b)}{a^{2}+ab+b^{2}}\geq a+b-\frac{2ab(a+b)}{3ab}\geq \frac{a+b}{3}$

$\Rightarrow \frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}\geq \frac{x^{3}+y^{3}}{3}$

$\Rightarrow \sum \frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}\geq \frac{2}{3}(x^{3}+y^{3}+z^{3})\geq 2xyz\geq 2$