Với x, y > 0 ta chứng minh:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\\ \Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\\ \Leftrightarrow\left(x+y\right)^2\ge4xy\\ \Leftrightarrow\left(x-y\right)^2\ge0(luônđúng)\)
Dấu "=" xảy ra khi x = y
Áp dụng vào bài toán:
\(\frac{1}{a+b+2c}=\frac{1}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\ \Rightarrow\frac{4ab}{a+b+2c}\le\frac{ab}{a+c}+\frac{ab}{b+c}\)
Tương tự: \(\frac{4bc}{b+c+2a}\le\frac{bc}{b+a}+\frac{bc}{c+a}\\ \frac{4ca}{c+a+2b}\le\frac{ca}{c+b}+\frac{ca}{a+b}\\ 4\left(\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\right)\le\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ca}{a+b}=b+a+c\left(dpcm\right)\)
Dấu "=" xảy ra khi a = b = c
