$\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2 $ Question -
Suppose that $a, b, c, d$ are four positive real numbers with sum 4.
Prove that
$$
\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2
$$
my doubt -
Solution.- According to AM-GM, we deduce that
$$
\begin{aligned}
\frac{a}{1+b^{2} c} &=a-\frac{a b^{2} c}{1+b^{2} c} \geq a-\frac{a b^{2} c}{2 b \sqrt{c}}=a-\frac{a b \sqrt{c}}{2} \\
&=a-\frac{b \sqrt{a \cdot a c}}{2} \geq a-\frac{b(a+a c)}{4}
\end{aligned}
$$
According to this estimation,
$$
\sum_{c y c} \frac{a}{1+b^{2} c} \geq \sum_{c y c} a-\frac{1}{4} \sum_{c y c} a b-\frac{1}{4} \sum_{c y c} a b c
$$
By AM-GM inequality again, it's easy to refer that
$$
\sum_{c y c} a b \leq \frac{1}{4}\left(\sum_{c y c} a\right)^{2}=4 \quad ; \quad \sum_{c y c} a b c \leq \frac{1}{16}\left(\sum_{c y c} a\right)^{3}=4
$$
now i did not understand how they got to this both results using am - gm
$$
\sum_{c y c} a b \leq \frac{1}{4}\left(\sum_{c y c} a\right)^{2}=4 \quad ; \quad \sum_{c y c} a b c \leq \frac{1}{16}\left(\sum_{c y c} a\right)^{3}=4
$$
i think i am missing something easy...
any hints ??
thankyou

$\sum_{cyc} ab = (a+c)(b+d) \leq \left(\frac{ (a+b+c+d)}{2} \right)^2$
This happens to work for 4 variables, but not other cases.

In Pham's book, example 1.1.4 is:  
$16 (abc+bcd+cda+dab) = 16ab(c+d)+16cd(a+b) $
$\leq 4(a+b)^2(c+d) + 4(c+d)^2 (a+b)$
$= 4 (a+b+c+d)(a+b)(c+d)$
$\leq (a+b+c+d)^3$.   
