a) Xét ΔABE  và ΔACF có

Alà góc chung

AEB=AFC(=90^O)

=> ΔABE đồng dạng ΔACF (g.g)

=>AF/AE​=AC/AB​

=> AB/AE​=AC/AF​

XétΔAEF và  ΔABC có

AB/AE​=AC/AF​

Và Agóc chung

Suy raΔAEF đồng dạngΔABC( c.g.c) 

\(A=\sqrt{x^3+8}+\sqrt{y^3+8}+\sqrt{z^3+8}\)

\(A=\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2x+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}\)

\(\sqrt{\frac{1}{2}}A=\sqrt{\left(x+2\right)\left(x^2-2x+4\right).\frac{1}{2}}+\sqrt{\left(y+2\right)\left(y^2-2x+4\right).\frac{1}{2}}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right).\frac{1}{2}}\)\(\sqrt{\frac{1}{2}}A=\sqrt{\left(x+2\right)\left(\frac{x^2}{2}-x+2\right)}+\sqrt{\left(y+2\right)\left(\frac{y^2}{2}-x+2\right)}+\sqrt{\left(z+2\right)\left(\frac{z^2}{2}-z+2\right)}\)

Áp dụng BĐT AM-GM ta có: 

\(\sqrt{\frac{1}{2}}A\le\frac{x+2+\frac{x^2}{2}-x+2+y+2+\frac{y^2}{2}-y+2+z+2+\frac{z^2}{2}-z+2}{2}=\frac{12+\frac{x^2+y^2+z^2}{2}}{2}=\frac{12+\frac{48}{2}}{2}=\frac{12+24}{2}=\frac{36}{2}=18\)

\(\Leftrightarrow A\le18:\sqrt{\frac{1}{2}}=18\sqrt{2}\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2=\frac{x^2}{2}-x+2\\y+2=\frac{y^2}{2}-y+2\\z+2=\frac{z^2}{2}-z+2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=4x\\y^2=4y\\z^2=4z\end{cases}}\Leftrightarrow\hept{\begin{cases}x\left(x-4\right)=0\\y\left(y-4\right)=0\\z\left(z-4\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=4\\z=4\end{cases}\left(v\text{ì}x,y,z>0\right)}}\)

Vậy \(A_{max}=18\sqrt{2}\Leftrightarrow x=y=z=4\)

Tham khảo nhé~

2k4 phải ko ??

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I don't know what are you say