Ko chịu tag@@
Ta có: \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=\frac{1}{3}\) (1)
Và \(xyz\le\frac{\left(x+y+z\right)^3}{27}=\frac{1}{27}\) (cô si cho 3 số)
\(\frac{1}{x^2+y^2+z^2}+\frac{1}{xyz}=\frac{1}{x^2+y^2+z^2}+\frac{1}{3xy}+\frac{1}{3yz}+\frac{1}{3zx}+\frac{1}{xyz}-\frac{1}{3}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\ge\frac{16}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}+\frac{1}{xyz}-\frac{1}{3}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\ge\frac{16}{1+\frac{1}{3}}+\frac{3}{3xyz}-\frac{x+y+z}{3xyz}\) (sử dụng (1) và quy đồng mấy cái phía sau)
\(=12+\frac{3-\left(x+y+z\right)}{3xyz}=12+\frac{2}{3xyz}\)
\(\ge12+\frac{2}{3.\frac{1}{27}}=30^{\left(đpcm\right)}\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
