Question

cho x>0; y>0 thỏa mãn  x+y=1. Tìm GTNN của biểu thức P=\(\left(1-\frac{1}{x^2}\right)\left(1-\frac{1}{y^2}\right)\)

Answer 1

Ta có: P = \(P=\left(1+\frac{1}{x}\right)\left(1-\frac{1}{y}\right).\left(1-\frac{1}{x}\right)\left(1-\frac{1}{y}\right)\) (HĐT số 3)
\(=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right).\frac{\left(x-1\right)\left(y-1\right)}{xy}\)

\(=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right).\frac{-x.-y}{xy}\)

= (1 + 1/x)(1 + 1/y) 
= 1 + 1/(xy) + (1/x + 1/y) = 1 + 1/(xy) + (x + y)/xy 
= 1 + 1/(xy) + 1/(xy) = 1 + 2/(xy) 
Áp dụng bđt: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)
 \(\Rightarrow P\ge\frac{1+2}{\frac{1}{4}}=9\) 
Vậy P_Min = 9 xảy ra \(\Leftrightarrow x=y=\) \(\frac{1}{2}\)