Do a,b > 0 => \(1-\frac{1}{a}\) và \(1-\frac{1}{b}\)luôn dương
Áp dụng bđt : \(xy\le\frac{\left(x+y\right)^2}{4}\) <=> \(\left(x+y\right)^2\ge4xy\) <=> \(\left(x-y\right)^2\ge0\) (luôn đúng)
P = \(\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\le\frac{1}{4}\left(1-\frac{1}{a}+1-\frac{1}{b}\right)^2=\frac{1}{4}\left[2-\left(\frac{1}{a}+\frac{1}{b}\right)\right]^2\)
Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) (a,b > 0) (1)
CM bđt đúng: Từ (1) <=> \(\left(\frac{x+y}{xy}\right)\left(x+y\right)\ge4\)
<=> \(\left(x+y\right)^2\ge4xy\) <=> \(\left(x-y\right)^2\ge0\) (luôn đúng)
Khi đó: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=\frac{4}{4}=1\)
=> \(2-\left(\frac{1}{a}+\frac{1}{b}\right)\le2-1=1\) => \(\frac{1}{4}\left[2-\left(\frac{1}{a}+\frac{1}{b}\right)\right]^2\le\frac{1}{4}.1^2=\frac{1}{4}\)
Dấu "=" xảy ra <=> a = b = 2
Vậy MaxP = 1/4 khi a =b = 2
