Question

Cho a,b,c>0 ;a+b+c=1

Tìm max A= \(\dfrac{ab}{\sqrt{c+ab}}\) +\(\dfrac{bc}{\sqrt{a+bc}}\)+\(\dfrac{ca}{\sqrt{b+ca}}\)

Answer 1

Lời giải:

Áp dụng BĐT AM-GM ngược dấu ta có:

\(A=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ac}}=\frac{ab}{\sqrt{c(a+b+c)+ab}}+\frac{bc}{\sqrt{a(a+b+c)+bc}}+\frac{ca}{\sqrt{b(a+b+c)+ac}}\)

\(=\frac{ab}{\sqrt{(c+a)(c+b)}}+\frac{bc}{\sqrt{(a+b)(a+c)}}+\frac{ca}{\sqrt{(b+a)(b+c)}}\)

\(\leq \frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}\right)+\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)+\frac{1}{2}\left(\frac{ca}{b+a}+\frac{ca}{b+c}\right)\)

\(A\leq \frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}(b+a+c)=\frac{1}{2}\)

Vậy \(A_{\max}=\frac{1}{2}\) tại \(a=b=c=\frac{1}{3}\)