 # I am stuck in the following olympiad problem: Suppose that a,b,c,d>=0 and a+b+c+d=4. Prove that a/(1+b^2c)+b/(1+c^2d)+c/(1+d^2a)+d/(1+a^2b)>=2 Kailey Vargas 2022-09-11 Answered
$\frac{a}{1+{b}^{2}c}+\frac{b}{1+{c}^{2}d}+\frac{c}{1+{d}^{2}a}+\frac{d}{1+{a}^{2}b}\ge 2$
I am stuck in the following olympiad problem:
Suppose that $a,b,c,d\ge 0$ and $a+b+c+d=4$. Prove that
$\frac{a}{1+{b}^{2}c}+\frac{b}{1+{c}^{2}d}+\frac{c}{1+{d}^{2}a}+\frac{d}{1+{a}^{2}b}\ge 2.$
Attempt. I tried to use reversed AM-GM technique.
$\begin{array}{rlrl}\frac{a}{1+{b}^{2}c}-a& =\frac{-a{b}^{2}c}{1+{b}^{2}c}& \phantom{\rule{2em}{0ex}}\frac{b}{1+{c}^{2}d}-b& =\frac{-b{c}^{2}d}{1+{c}^{2}d}\\ \frac{c}{1+{d}^{2}a}-c& =\frac{-c{d}^{2}a}{1+{d}^{2}a}& \phantom{\rule{2em}{0ex}}\frac{d}{1+{a}^{2}b}-d& =\frac{-d{a}^{2}b}{1+{a}^{2}b}\end{array}$
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By C-S and AM-GM we obtain:
$\sum _{cyc}\frac{d}{1+{a}^{2}b}=\sum _{cyc}\frac{{d}^{2}}{d+{a}^{2}bd}\ge \frac{\left(a+b+c+d{\right)}^{2}}{\sum _{cyc}\left(d+{a}^{2}bd\right)}=$
$=\frac{16}{4+\left(ab+cd\right)\left(ad+bc\right)}\ge \frac{16}{4+{\left(\frac{ab+bc+cd+da}{2}\right)}^{2}}=$
$=\frac{16}{4+{\left(\frac{\left(a+c\right)\left(b+d\right)}{2}\right)}^{2}}\ge \frac{16}{4+{\left(\frac{{\left(\frac{a+c+b+d}{2}\right)}^{2}}{2}\right)}^{2}}=2.$

We have step-by-step solutions for your answer! themediamafia73
$\sum _{cyc}\frac{d}{1+{a}^{2}b}=\sum _{cyc}\frac{{d}^{2}}{d+{a}^{2}bd}$
Hence by Titu's lemma
$\sum _{cyc}\frac{{d}^{2}}{d+{a}^{2}bd}\ge \frac{\left(a+b+c+d{\right)}^{2}}{\sum _{cyc}\left(d+{a}^{2}bd\right)}=$
$=\frac{16}{4+\left(ab+cd\right)\left(ad+bc\right)}$
Now by AM GM inequality we have
$\frac{ab+bc+cd+ad}{2}\ge \sqrt{\left(ab+cd\right)\left(ad+bc\right)}$
Hence we get
${\left(\frac{ab+bc+cd+ad}{2}\right)}^{2}\ge \left(ab+cd\right)\left(ad+bc\right)$
$⇒\frac{16}{4+\left(ab+cd\right)\left(ad+bc\right)}\ge \frac{16}{4+{\left(\frac{ab+bc+cd+ad}{2}\right)}^{2}}$
Now again by AM GM inequality
$\frac{a+b+c+d}{2}\ge \sqrt{\left(a+c\right)\left(d+b\right)}$
$⇒\frac{16}{4+{\left(\frac{\left(a+c\right)\left(b+d\right)}{2}\right)}^{2}}\ge \frac{16}{4+{\left(\frac{{\left(\frac{a+b+c+d}{2}\right)}^{2}}{2}\right)}^{2}}$
$=2.$

We have step-by-step solutions for your answer!