Ta có: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)
\(>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)
\(=\frac{a+b+c+d}{a+b+c+d}=1\)
Tương tự ta cũng chứng minh được \(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1\)
mà \(\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\right)+\left(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}\right)\)
\(=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{d+a}=4\)
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)là số nguyên
do đó \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\)
\(\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b}{a+b}-\frac{b}{b+c}+\frac{d}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)(vì \(a\ne c\))
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac=bd\)(vì \(b\ne d\))
Khi đó \(abcd=ac.ac=\left(ac\right)^2\)là số chính phương.
