$\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

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

Answer 1

$\sum_{cyc} ab = (a+c)(b+d) \leq \left(\frac{ (a+b+c+d)}{2} \right)^2$

This happens to work for 4 variables, but not other cases.

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In Pham's book, example 1.1.4 is:

$16 (abc+bcd+cda+dab) = 16ab(c+d)+16cd(a+b) $
$\leq 4(a+b)^2(c+d) + 4(c+d)^2 (a+b)$
$= 4 (a+b+c+d)(a+b)(c+d)$
$\leq (a+b+c+d)^3$.