Problem: If $abcde = 1$, $a, b, c, d, e > 0$, $a, b, c, d, e \in \Bbb R$, prove that
$\sum_{cyc} {{a+abc} \over {1+ab+abcd}} \ge {{10} \over {3}}$
First I proceeded with Cauchy-Schwartz inequality, but I couldn't find any way to use that. I think that the equal condition is $a=b=c=d=e=1$.
I don't know what to do to solve this problem. Can somebody please give me a hint?
Yes, C-S helps!
Indeed, let $a=\frac{y}{z},$ $b=\frac{z}{y}$, $c=\frac{t}{z}$, $d=\frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.
Thus, the condition gives $e=\frac{x}{w}$ and we obtain: $$\sum_{cyc}\frac{a+abc}{1+ab+abcd}=\sum_{cyc}\frac{\frac{y}{x}+\frac{y}{x}\cdot\frac{z}{y}\cdot\frac{t}{z}}{1+\frac{y}{x}\cdot\frac{z}{y}+\frac{y}{x}\cdot\frac{z}{y}\cdot\frac{t}{z}\cdot\frac{w}{t}}=$$ $$=\sum_{cyc}\frac{\frac{y}{x}+\frac{t}{x}}{1+\frac{z}{x}+\frac{w}{x}}=\sum_{cyc}\frac{y+t}{x+z+w}=-5+\sum_{cyc}\frac{x+y+z+t+w}{x+z+w}=$$ $$=-5+\frac{1}{3}\sum_{cyc}(x+z+w)\sum_{cyc}\frac{1}{x+z+w}.$$ Can you end it now?