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Ta có:
\(A=\left(x^2+\frac{1}{8x}+\frac{1}{8x}\right)+\left(y^2+\frac{1}{8y}+\frac{1}{8y}\right)+\left(z^2+\frac{1}{8z}+\frac{1}{8z}\right)+\frac{6}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}+3\sqrt[3]{y^2.\frac{1}{8y}.\frac{1}{8y}}+3\sqrt[3]{z^2.\frac{1}{8z}.\frac{1}{8z}}+\frac{6}{8}\frac{9}{x+y+z}\)
\(=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{6}{8}.\frac{9}{\frac{3}{2}}=\frac{27}{4}\)
Dấu "=" xảy ra <=> x = y = z = 1/2
Vậy min A = 27/4 tại x = y = z = 1/2
2. ĐK: \(x\ge-5\)
\(\Leftrightarrow\left(x+5-6\sqrt{x+5}+9\right)+\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+5}-3\right)^2+\left(x-4\right)^2=0\)
\(\forall x\ge-5\) ta luôn có \(\left(\sqrt{x+5}-3\right)^2+\left(x-4\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\sqrt{x+5}-3=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) x = 4 (nhận)
\(M\left(x+y+z\right)=\left(z^2+y^2+z^2\right)+2+\frac{\left(x^2+1\right)\left(y+z\right)}{x}+\frac{\left(y^2+1\right)\left(z+x\right)}{y}+\frac{\left(z^2+1\right)\left(x+y\right)}{z}\)
\(=5+\frac{\left(x^2+1\right)\left(y+z\right)}{x}+\frac{\left(y^2+1\right)\left(z+x\right)}{y}+\frac{\left(z^2+1\right)\left(x+y\right)}{z}\)
\(\ge5+2\left(y+z\right)+2\left(z+x\right)+2\left(x+y\right)=5+4\left(x+y+z\right)\) ( Sử dụng BĐT Cô-si cho 2 số dương ý)
\(\Rightarrow M\ge\frac{5}{x+y+z}+4\)
Mặt khác: \(\left(x+y+z\right)^2\le\left(x^2+y^2+z^2\right)\left(1^2+1^2+1^2\right)=9\)
\(\Rightarrow x+y+z\le3\)
Do đó: \(M\ge\frac{5}{3}+4=\frac{17}{3}\)
\(M=\frac{17}{3}\Leftrightarrow x=y=z=1\)
\(\Rightarrow Min_A=\frac{17}{3}\)
\(A\ge\frac{\left(x+y+z\right)^2}{3}+\frac{9}{x+y+z}=\frac{\left(x+y+z\right)^2}{3}+\frac{9}{8\left(x+y+z\right)}+\frac{9}{8\left(x+y+z\right)}+\frac{27}{4\left(x+y+z\right)}\)
\(A\ge3\sqrt[3]{\frac{81\left(x+y+z\right)^2}{3.64\left(x+y+z\right)\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{27}{4}\)
\(A_{min}=\frac{27}{4}\) khi \(x=y=z=\frac{1}{2}\)