Question

Chứng minh với mọi số \(a,b,c,d>0\) thì \(\dfrac{a-d}{b+d}+\dfrac{d-b}{b+c}+\dfrac{b-c}{c+a}+\dfrac{c-a}{a+d}\ge0\)

Answer 1

\(\Leftrightarrow\dfrac{a-d}{b+d}+1+\dfrac{d-b}{b+c}+1+\dfrac{b-c}{c+a}+1+\dfrac{c-a}{a+d}+1\ge4\)

\(\Leftrightarrow\dfrac{a+b}{b+d}+\dfrac{c+d}{b+c}+\dfrac{a+b}{c+a}+\dfrac{c+d}{a+d}\ge4\)

\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{b+d}+\dfrac{1}{c+a}\right)+\left(c+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)\ge4\)

BĐT nói trên đúng do:

\(\dfrac{1}{b+d}+\dfrac{1}{c+a}\ge\dfrac{4}{a+b+c+d}\) và \(\dfrac{1}{b+c}+\dfrac{1}{a+d}\ge\dfrac{4}{a+b+c+d}\)