Question

HSG Phú Thọ 2016

Cho các số dương a,b,c thỏa mãn \(ab+bc+ca=1\). Chứng minh rằng :

       \(\frac{a-b}{1+c^2}+\frac{b-c}{1+a^2}+\frac{c-a}{1+b^2}=0\)

Answer 1

Ta có 1+c²=ab+bc+ca+c²=(a+c)(b+c)

Tương tự 1+a²=(a+b)(a+c)

                 1+b²=(a+b)(b+c)

Suy ra \(\frac{a-b}{1+c^2}=\frac{a-b}{\left(a+c\right)\left(b+c\right)}=\frac{1}{c+b}-\frac{1}{c+a}\)

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

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

\(\Rightarrow\frac{a-b}{1+c^2}+\frac{b-c}{1+a^2}+\frac{c-a}{1+b^2}=\frac{1}{c+b}-\frac{1}{c+a}+\frac{1}{a+c}-\frac{1}{a+b}+\frac{1}{a+b}-\frac{1}{b+c}=0\)