Bài làm
\(P=2a+3b+\frac{4}{a}+\frac{9}{b}=a+a+2b+b+\frac{4}{a}+\frac{9}{b}\)
\(=\left(a+2b\right)+\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)\)
\(\ge8+2\sqrt{a\times\frac{4}{a}}+2\sqrt{b\times\frac{9}{b}}\)( Cauchy )
\(=8+4+6=18\)
Đẳng thức xảy ra khi a = 2 ; b = 3
=> MinP = 18 <=> a = 2 ; b = 3
\(P=2a+3b+\frac{4}{a}+\frac{9}{b}\)
\(\Leftrightarrow P=\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)+a+2b\)
Áp dụng BĐT AM-GM ta có:
\(P\ge2.\sqrt{a.\frac{4}{a}}+2.\sqrt{b.\frac{9}{b}}+a+2b=2.2+2.3+a+2b\ge4+6+8=18\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a=\frac{4}{a}\\b=\frac{9}{b}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
Vậy \(P_{min}=18\)\(\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
\(P=a+2b+a+\frac{4}{a}+b+\frac{9}{b}\)
Áp dụng cô si ta có:
\(a+\frac{4}{a}\ge4\)
\(b+\frac{9}{b}\ge6\)
\(\Rightarrow P\ge4+6+8\Rightarrow P\ge18\)
dấu "=" xảy ra khi a=2,b=3
