Có P\(=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+1+1+\frac{1}{x^2y^2}=2+x^2y^2+\frac{1}{x^2y^2}=\left(xy+\frac{1}{xy}\right)^2\)
Áp dụng bđt cosi vs hai số dương có:
\(xy+\frac{1}{16xy}\ge2\sqrt{xy.\frac{1}{16xy}}=2\sqrt{\frac{1}{16}}=\frac{1}{2}\)(1)
Có \(\left(x+y\right)^2\ge4xy\) với mọi x,y
<=> \(1\ge4xy\) (do x+y=1)
=> \(\frac{1}{xy}\ge4\) <=> \(\frac{15}{16xy}\ge\frac{15}{16}.4=\frac{15}{4}\)(2)
Cộng vế vs vế của (1),(2) có \(xy+\frac{1}{16xy}+\frac{15}{16xy}\ge\frac{1}{2}+\frac{15}{4}\)
<=> \(xy+\frac{1}{xy}\ge\frac{17}{4}\)
<=> \(\left(xy+\frac{1}{xy}\right)^2\ge\frac{289}{16}\)
<=> \(P\ge\frac{289}{16}\)
Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)
