# Math Help - Fraction Chain Inequality Proof

1. ## Fraction Chain Inequality Proof

Prove that

$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d} {e+f}+\frac{e}{f+a}+\frac{f}{a+b}\geq 3$

Condition: a, b, c, d, e, f are positive real numbers.

2. 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:

$
\frac{a}{b+c}+\frac{b}{c+d}
$

Add the two by multiplying both with the denominator of the other.

$(b+c)*(c+d) = bc + cd + c^2 + bd$

$
\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

$
\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: $3 + \frac{...}{...}$ and because all the numbers are positive and real, this fraction will never produce a negative number.

3. Originally Posted by Pim
bc occurs both in the numerator and denominator, so this simplifies to

$
\frac{ac + ad + bc + b^2}{bc + cd + c^2 + bd} = 1 + \frac{ac + ad + b^2}{ cd + c^2 + bd}
$
Who told you?

Anyway, the inequality is the Shapiro inequality for $n=6$. You might find a proof along the same lines as those given for the simpler Nesbitt’s inequality on the Wikipedia page.

4. Originally Posted by JaneBennet
Who told you?

Anyway, the inequality is similar to Nesbitt’s inequality. You might find a proof along the same lines as those given on the Wikipedia page.
You can't cancel the $bc$ terms like that, right?

And how does the Nesbitt Inequality extend to this 6-term inequality?