Question

[Pham Van Hung]

Cho a;b;c > 0 thỏa mãn a + b + c = 3

Tìm GTLN của \(D=\frac{ab}{\sqrt{c^2+3}}+\frac{bc}{\sqrt{a^2+3}}+\frac{ca}{\sqrt{b^2+3}}\)

Answer 1

[tth_new]

Có: \(9=\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow3\ge ab+bc+ca\)

Từ đây: \(D=\Sigma_{cyc}\frac{ab}{\sqrt{c^2+3}}\le\Sigma_{cyc}\frac{ab}{\sqrt{c^2+ab+bc+ca}}\)

\(=\Sigma_{cyc}\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=\Sigma_{cyc}\sqrt{\frac{ab}{a+c}}.\sqrt{\frac{ab}{b+c}}\le\Sigma_{cyc}\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

\(=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)

Đẳng thức xảy ra khi a = b = c = 1