Question

cho a,b,c là 3 số thực dương và có tổng bằng 1. Chứng minh:\(\dfrac{a-bc}{a+bc}+\dfrac{b-ac}{b+ac}+\dfrac{c-ab}{c+ab}\le\dfrac{3}{2}\)

Answer 1

Đặt \(P=\dfrac{a-bc}{a+bc}+\dfrac{b-ac}{b+ac}+\dfrac{c-ab}{c+ab}\)

\(P=\dfrac{a-bc}{a+bc}-1+\dfrac{b-ac}{b+ac}-1+\dfrac{c-ab}{c+ab}-1+3\)

\(=3-2\left(\dfrac{bc}{a+bc}+\dfrac{ac}{b+ac}+\dfrac{ab}{c+ab}\right)\)

\(=3-2\left(\dfrac{a^2b^2}{abc+a^2b^2}+\dfrac{b^2c^2}{abc+b^2c^2}+\dfrac{c^2a^2}{abc+c^2a^2}\right)\)

\(\le3-\dfrac{2\left(ab+bc+ca\right)^2}{a^2b^2+b^2c^2+c^2a^2+3abc}=3-\dfrac{2\left(ab+bc+ca\right)^2}{a^2b^2+b^2c^2+c^2a^2+3abc\left(a+b+c\right)}\)

\(=3-\dfrac{2\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2+abc\left(a+b+c\right)}\le3-\dfrac{2\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2+\dfrac{1}{3}\left(ab+bc+ca\right)^2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)