Question

[Nguyễn Thu Ngà]

cho a,b dương thỏa mãn a+b+2ab=12

Tìm min A=\(\dfrac{a^2+ab}{a+2b}+\dfrac{b^2+ab}{2a+b}\)

Answer 1

[Akai Haruma]

Lời giải:

\(A=\frac{a(a+b)}{a+2b}+\frac{b(b+a)}{2a+b}=(a+b)\left(\frac{a}{a+2b}+\frac{b}{2a+b}\right)\)

Áp dụng BĐT Cauchy_Schwarz và AM-GM:

\(\frac{a}{a+2b}+\frac{b}{2a+b}=\frac{a^2}{a^2+2ab}+\frac{b^2}{2ab+b^2}\geq \frac{(a+b)^2}{(a+b)^2+2ab}\geq \frac{(a+b)^2}{(a+b)^2+\frac{(a+b)^2}{2}}=\frac{2}{3}\)

Do đó:

\(A\geq \frac{2(a+b)}{3}\)

Cũng theo BĐT AM-GM: \(12=a+b+2ab\leq a+b+\frac{(a+b)^2}{2}\)

\(\Leftrightarrow (a+b)^2+2(a+b)-24\geq 0\)

\(\Leftrightarrow (a+b-4)(a+b+6)\geq 0\Rightarrow a+b\geq 4\)

\(\Rightarrow A\geq \frac{2}{3}(a+b)\geq \frac{8}{3}\)

Vậy \(A_{\min}=\frac{8}{3}\Leftrightarrow a=b=2\)