Question

Answer 1

-Ta có BĐT \(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\) với 3 số dương x,y,z.

\(\Rightarrow xy+yz+zx\le\dfrac{1^2}{3}=\dfrac{1}{3}\)

\(\left(xy+yz+zx\right)=\dfrac{1}{3}\Leftrightarrow x=y=z=\dfrac{1}{3}\).

-Vậy \(\left(xy+yz+zx\right)_{max}=\dfrac{1}{3}\)

b) Áp dụng BĐT Caushy Schwarz ta có:

\(P=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{\left(1+1+1\right)^2}{xy+yz+zx}\ge\dfrac{9}{\dfrac{1}{3}}=27\)

\(P=27\Leftrightarrow x=y=z=\dfrac{1}{3}\)

-Vậy \(P_{min}=27\)