a) Áp dụng Bất đẳng thức Cauchy
\(P=\left(1-\dfrac{1}{a}\right)\left(1-\dfrac{1}{b}\right)\le\dfrac{1}{4}\left(1-\dfrac{1}{a}+1-\dfrac{1}{b}\right)^2=\dfrac{1}{4}\left[2-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]^2\)
mà \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}=1\left(a+b=4\right)\)
\(\Rightarrow P\le\dfrac{1}{4}\left(2-1\right)^2=\dfrac{1}{4}\)
\(\Rightarrow GTLN\left(P\right)=\dfrac{1}{4}\left(tại.a=b=2\right)\)
b) Áp dụng Bất đẳng thức Cauchy
\(F=\left(1+\dfrac{1}{a}\right)^2+\left(1+\dfrac{1}{b}\right)^2\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+1+\dfrac{1}{b}\right)\)
\(\Rightarrow F\ge\dfrac{1}{2}\left(2+\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)
mà \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}=4\left(a+b=1\right)\)
\(\Rightarrow F\ge\dfrac{1}{2}\left(2+4\right)^2=18\)
\(\Rightarrow GTNN\left(F\right)=18\left(tại.a=b=\dfrac{1}{2}\right)\)
