Ta có : \(a;b>0,3a+2b\le6\Rightarrow\dfrac{a}{2}+\dfrac{b}{3}\le1\) .
Dự đoán : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{1}{2}\Rightarrow a=1;b=\dfrac{3}{2}\)
Đặt : x = a/2 ; y = b/3 ( x ; y > 0 ) ; ta có : \(x+y\le1\)
\(P=\left(1-\dfrac{4}{a^2}\right)\left(1-\dfrac{9}{b^2}\right)=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)\) = \(\dfrac{\left(x^2-1\right)\left(y^2-1\right)}{x^2y^2}\)
\(\left(x^2-1\right)\left(y^2-1\right)=\left(1-x^2\right)\left(1-y^2\right)=\left(1+x\right)\left(1+y\right)\left(1-x\right)\left(1-y\right)\ge\left(1+x\right)\left(1+y\right)yx\)
Suy ra : \(P\ge\dfrac{xy\left(1+x\right)\left(1+y\right)}{x^2y^2}=\dfrac{xy+x+y+1}{xy}=1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}=Q\)
Thật vậy ; ta có : \(Q\ge1+\dfrac{4}{x+y}+\dfrac{4}{\left(x+y\right)^2}\ge1+4+4=9\)
Suy ra : \(P\ge9\) . " = " \(\Leftrightarrow x=y=\dfrac{1}{2}\Leftrightarrow a=1;b=\dfrac{3}{2}\)
