\(VT=x^2+y^2+z^2+\frac{49}{x+2y+3z}\)
\(=\frac{x^2}{1}+\frac{\left(2y\right)^2}{4}+\frac{\left(3y\right)^2}{9}+\frac{49}{\left(x+2y+3z\right)}\)
\(\ge\frac{\left(x+2y+3z\right)^2}{14}+\frac{49}{2\left(x+2y+3z\right)}+\frac{49}{2\left(x+2y+3z\right)}\)
\(\ge3\sqrt[3]{\frac{\left(x+2y+3z\right)^2}{14}.\frac{49}{2\left(x+2y+3z\right)}.\frac{49}{2\left(x+2y+3z\right)}}=\frac{21}{2}\)
Đẳng thức xảy ra khi \(x=\frac{1}{2};y=1;z=\frac{3}{2}\)
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