An algebra problem by Baibhab Chakraborty

From Titu's Lemma ,

$S = \frac{a}{1 + b^{2}c} + \frac{b}{1 + c^{2}d} + \frac{c}{1 + d^{2}a} + \frac{d}{1 + a^{2}b}$

= $\frac{a^{2}}{a + ab^{2}c} + \frac{b^{2}}{b + bc^{2}d} + \frac{c^{2}}{c + cd^{2}a} + \frac{d^{2}}{d + da^{2}b}$

$\ge \frac{{(a + b + c + d)}^{2}}{a + b + c + d + abcd( \frac{b}{d} + \frac{d}{b} + \frac{c}{a} + \frac{a}{c})}$

$\ge \frac{ 16 }{ 4 + abcd( \frac{b}{d} + \frac{d}{b} + \frac{a}{c}+ \frac{c}{a} ) }$

Now

$\frac{a}{c} + \frac{c}{a} + \frac{b}{d} + \frac{d}{b} \ge 4$ ,

and $abcd \le ( \frac{a + b + c + d}{4} )^{4}$ or $abcd \le 1$

both of which are obtained from AM.GM.Inequality .

So, by multiplying ,

$abcd( \frac{a}{c} + \frac{c}{a} + \frac{b}{d} + \frac{d}{b} ) \le 4$

$4 + abcd( \frac{a}{c} + \frac{c}{a} + \frac{b}{d} + \frac{d}{b} ) \le 8$

$\frac{1}{4 + abcd( \frac{a}{c} + \frac{c}{a} + \frac{b}{d} + \frac{d}{b}} ) \ge \frac{1}{8}$

$S \ge 2$ .

Hope this helps. :)