Cần chứng minh \(\frac{a-d}{b+d}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\ge0\)
Ta có \(\frac{a-d}{b+d}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}=\frac{\left(a+b\right)-\left(b+d\right)}{b+d}+\frac{\left(c+d\right)-\left(b+c\right)}{b+c}+\frac{\left(a+b\right)-\left(c+a\right)}{c+a}+\frac{\left(c+d\right)-\left(a+d\right)}{a+d}\)\(=\frac{a+b}{b+d}-1+\frac{c+d}{b+c}-1+\frac{a+b}{c+a}-1+\frac{c+d}{a+d}-1\)
\(=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{c+a}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)
Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) được :
\(\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{c+a}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\ge\frac{4\left(a+b\right)}{a+b+c+d}+\frac{4\left(c+d\right)}{a+b+c+d}-4\)\(=\frac{4\left(a+b+c+d\right)}{a+b+c+d}-4=4-4=0\)
Suy ra ta có điều phải chứng minh.
