Tìm min và max

E

eye_smile

Ta có: $\dfrac{x}{1-yz} \le \dfrac{x}{1-\dfrac{y^2+z^2}{2}}=\dfrac{2x}{x^2+1}$

Tương tự, có: $\dfrac{y}{1-xz} \le \dfrac{2y}{y^2+1}$

$\dfrac{z}{1-xy} \le \dfrac{2z}{z^2+1}$

\Rightarrow $A=\dfrac{x}{1-yz}+\dfrac{y}{1-zx}+\dfrac{z}{1-xy} \le \dfrac{2x}{x^2+1}+\dfrac{2y}{y^2+1}+\dfrac{2z}{z^2+1}$

Đặt $x^2+1=a;y^2+1=b;z^2+1=c$ \Rightarrow $a+b+c=4$

\Rightarrow $A \le \dfrac{2\sqrt{a-1}}{a}+\dfrac{2\sqrt{b-1}}{b}+\dfrac{2\sqrt{c-1}}{c} \le \dfrac{3a-2}{\sqrt{3}a}+\dfrac{3b-2}{\sqrt{3}b}+ \dfrac{3c-2}{\sqrt{3}c}=3\sqrt{3}-\dfrac{2}{\sqrt{3}}(\dfrac{1}{a}+\dfrac{1}{b}+ \dfrac{1}{c}) \le 3\sqrt{3}-\dfrac{2}{\sqrt{3}}.\dfrac{9}{4}=\dfrac{3\sqrt{3}}{2}$

Dấu "=" xảy ra \Leftrightarrow $x=y=z=\dfrac{1}{\sqrt{3}}$
 
You must log in or register to reply here.
Top Bottom