Prove that
$\displaystyle \frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d} {e+f}+\frac{e}{f+a}+\frac{f}{a+b}\geq 3$
Condition: a, b, c, d, e, f are positive real numbers.
Edit: I'm not sure how awake I was when I typed this. This is to remain here purely to remind me of my own stupidity.
Take the first part:
$\displaystyle
\frac{a}{b+c}+\frac{b}{c+d}
$
Add the two by multiplying both with the denominator of the other.
$\displaystyle (b+c)*(c+d) = bc + cd + c^2 + bd$
$\displaystyle
\frac{ac + ad}{bc + cd + c^2 + bd} + \frac{bc + b^2}{bc + cd + c^2 + bd} = \frac{ac + ad + bc + b^2}{bc + cd + c^2 + bd}
$
bc occurs both in the numerator and denominator, so this simplifies to
$\displaystyle
\frac{ac + ad + bc + b^2}{bc + cd + c^2 + bd} = 1 + \frac{ac + ad + b^2}{ cd + c^2 + bd}
$
You can do this with the third and forth as well as the fifth and sixth fractions yielding an expression in the form of: $\displaystyle 3 + \frac{...}{...} $ and because all the numbers are positive and real, this fraction will never produce a negative number.
Who told you?
Anyway, the inequality is the Shapiro inequality for $\displaystyle n=6$. You might find a proof along the same lines as those given for the simpler Nesbitt’s inequality on the Wikipedia page.