Tìm GTLN biết x>0;y>0;z>0; x+y+z=1 A= $\sqrt[]{x²+xyz}$ + $\sqrt[]{y²+xyz}$ + $\sqrt[]{z²+xyz}$ + 9.$\sqrt[]{xyz}$

Đáp án:

Giải thích các bước giải:

 Ta có:

$\sqrt{x^2+xyz}=\dfrac{\sqrt{3}}{2}\sqrt{\dfrac{4x}{3}(x+yz)} \leq \dfrac{\sqrt{3}}{4}\left(\dfrac{4x}{3}+x+yz \right)$

$⇔\sqrt{x^2+xyz} \leq \dfrac{\sqrt{3}}{4}\left(\dfrac{7x}{3}+yz \right)$

Hoàn toàn tương tự: 

$\sqrt{y^2+xyz} \leq \dfrac{\sqrt{3}}{4}\left(\dfrac{7y}{3}+zx \right)$

$\sqrt{z^2+xyz} \leq \dfrac{\sqrt{3}}{4}\left(\dfrac{7z}{3}+xy \right)$

Cộng vế với vế:

$⇒A \leq \dfrac{\sqrt{3}}{4}\left(\dfrac{7x}{3}+\dfrac{7y}{3}+\dfrac{7z}{3}+xy+yz+zx \right)+9\sqrt{xyz}$

$⇒A \leq \dfrac{7\sqrt{3}}{12}+\dfrac{\sqrt{3}}{4}\left(xy+yz+zx \right)+9\sqrt{xyz}$

Ta lại có:

$xy+yz+zx \leq \dfrac{1}{3}(x+y+z)^2=\dfrac{1}{3}$

Và 

$xyz \leq \dfrac{1}{27}(x+y+z)^3=\dfrac{1}{27}⇒9\sqrt{xyz} \leq \sqrt{3}$

Do đó:

$A \leq \dfrac{7\sqrt{3}}{12}+\dfrac{\sqrt{3}}{4}·\dfrac{1}{3}+\sqrt{3}=\dfrac{5\sqrt{3}}{3}$

Vậy $A_{max}=\dfrac{5\sqrt{3}}{3}$ khi $x=y=z=1$