\(P=45a+20b+\dfrac{27}{b^2}+\dfrac{108}{a}\)
\(P=27\left(a+\dfrac{4}{a}\right)+\left(b+b+\dfrac{27}{b^2}\right)+18\left(a+b\right)\)
\(P\ge27.2\sqrt{\dfrac{4a}{a}}+3\sqrt[3]{\dfrac{27b^2}{b^2}}+18.5=207\)
Dấu "=" xảy ra khi \(\left(a;b\right)=\left(2;3\right)\)
\(45a+20b+\frac{27\left(a+4b^2\right)}{ab^2}=45a+20b+\frac{27}{b^2}+\frac{108}{a}\)
\(=\left(\frac{108}{a}+27a\right)+\left(\frac{27}{b^2}+b+b\right)+18\left(a+b\right)\)
\(\ge108+9+18\cdot5=207\)
tks mấy bạn ngu quá ko nhìn ra nổi ToT
nhma cho hỏi cái là làm sao lên dc \(b+b+\frac{27}{b^2}\ge3\sqrt[3]{\frac{27b^2}{b^2}}\) vậy? ;-;