If,a,b,c,d>0 ,and <br />
abcd\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \right) = a + b + c + d
Ten demonstrate that:
<br />
\frac{{\left( {a + 1} \right)^2 }}{{a + b}} + \frac{{\left( {b + 1} \right)^2 }}{{b + c}} + \frac{{\left( {c + 1} \right)^2 }}{{c + d}} + \frac{{\left( {d + 1} \right)^2 }}{{d + a}} \le 2(a + b + c + d)<br />