Question

cho  x,  y >0 và  x+y\(\le\)2 tìm  gtnn của  bt

\(p=\frac{20}{x^2+y^2}+\frac{11}{xy}\)

Answer 1

Từ BĐT \(\left(x+y\right)^2\ge4xy\) ta suy ra \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) và \(\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)

Ta có : \(P=\frac{20}{x^2+y^2}+\frac{11}{xy}=20\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{xy}\ge20.\frac{4}{\left(x+y\right)^2}+\frac{4}{\left(x+y\right)^2}\ge\frac{80}{4}+\frac{4}{4}=21\)

Dấu "=" xảy ra khi x = y = 1

Vậy Min P = 21 khi x = y = 1

Answer 2

Ta có :

\(P=\frac{20}{x^2+y^2}+\frac{11}{xy}\)

\(=20.\left[\frac{1}{x^2+y^2}+\frac{1}{2xy}\right]+\frac{1}{xy}\)

\(\ge20\cdot\frac{4}{x^2+y^2+2xy}+\frac{4}{\left(x+y\right)^2}\)

\(\ge20\cdot\frac{4}{2^2}+\frac{4}{2^2}=21\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=1\)

Vậy \(P_{min}=21\) khi \(x=y=1\)