* Với a, b, c > 0 ta có:
\(A=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)\(=\left(\frac{a}{b+c}+\frac{c}{d+a}\right)+\left(\frac{b}{c+d}+\frac{d}{a+b}\right)\)
\(=\)\(\frac{a\left(a+d\right)+c\left(b+c\right)}{\left(b+c\right)\left(d+a\right)}+\frac{b\left(a+b\right)+d\left(c+d\right)}{\left(a+b\right)\left(c+d\right)}\)\(\ge\frac{a^2+c^2+ad+bc}{\frac{1}{4}\left(a+b+c+d\right)^2}+\frac{b^2+d^2+ab+cd}{\frac{1}{4}\left(a+b+c+d\right)^2}\)\(=\frac{4\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)}{\left(a+b+c+d\right)^2}\) (Theo bất đẳng thức \(xy\le\frac{1}{4}\left(x+y\right)\))
Mặt khác:
\(2\left(a^2+b^2+c^2+d^2+ab+ad+bc+cd\right)-\left(a+b+c+d\right)^2\)
\(=a^2+b^2+c^2+d^2-2ac-2ad=\left(a-c\right)^2+\left(b-d\right)^2\ge0\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=c\\b=d\end{matrix}\right.\)
* Áp dụng: \(\frac{2016}{x+y}+\frac{x}{y+2016}+\frac{y}{4031}+\frac{2015}{x+2016}=2\)
\(\Rightarrow\)\(x=2015\), \(y=2016\)
