UCT nào
Ta chứng minh rằng: \(\dfrac{1}{a}+a+1\ge\dfrac{3}{4}a+2\)
Thật vậy, ta có: \(\dfrac{1}{a}+a+1=\dfrac{3}{4}a+\dfrac{1}{4}a+\dfrac{1}{a}+1\ge\dfrac{3}{4}a+2\sqrt{\dfrac{1}{4}a.\dfrac{1}{a}}+1=\dfrac{3}{4}a+2\)
\(\Rightarrow\left(\dfrac{1}{a}+a+1\right)^3\ge\left(\dfrac{3}{4}a+2\right)^3\)
Tương tự: \(\left(\dfrac{1}{b}+b+1\right)^3\ge\left(\dfrac{3}{4}b+2\right)^3\)
Cộng vế theo vế, áp dụng AM-GMta được:
\(P\ge\left(\dfrac{3}{4}a+2\right)^3+\left(\dfrac{3}{4}b+2\right)^3=\left(\dfrac{3}{4}a+2+\dfrac{3}{4}b+2\right)-3\left(\dfrac{3}{4}a+2\right)\left(\dfrac{3}{4}b+2\right)\left(\dfrac{3}{4}a+2+\dfrac{3}{4}b+2\right)\)
\(P\ge\left[\dfrac{3}{4}\left(a+b\right)+4\right]^3-3.\dfrac{\left(\dfrac{3}{4}a+2+\dfrac{3}{4}b+2\right)^2}{4}.\left[\dfrac{3}{4}\left(a+b\right)+4\right]=85,75\)
GTNN của P là 85,75 khi a=b=2
