2 replies to this topic

[xuantungjinkaido]

Cho x+y+z=0  $x^{2}+y^{2}+z^{2}=1$ Tìm max min $Q=x^{3}+y^{3}+z^{3}$

Edited by tpdtthltvp, 15-05-2016 - 22:58.

[Ngoc Hung]

Vì $x+y+z=0\Rightarrow Q=x^{3}+y^{3}+z^{3}=3xyz$

Áp dụng CauChy ta có

$1=x^{2}+y^{2}+z^{2}\geq 3\sqrt[3]{(xyz)^{2}}\Rightarrow (xyz)^{2}\leq \frac{1}{27}\Rightarrow -\frac{1}{\sqrt{3}}\leq Q\leq \frac{1}{\sqrt{3}}$

[thank you]

bác cho cháu hoi luon neu Q=x^5+y^5+z^5 thi lam sao được không ạ.