Nesbitt's inequality

Remarks

Proof idea: let $s = a + b + c$. Adding $1$ to each fraction gives $\frac{a}{b+c} + 1 = \frac{s}{b+c}$, so the left side equals $s \big( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \big) - 3$. By the Cauchy–Schwarz inequality, or the inequality between arithmetic and harmonic means, $\big((b+c) + (c+a) + (a+b)\big)\big(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\big) \ge 9$. Since the first factor is $2s$, the left side is at least $\frac{9}{2} - 3 = \frac{3}{2}$.

Example: $a = b = 1$, $c = 2$ gives $\frac{1}{3} + \frac{1}{3} + 1 = \frac{5}{3} \ge \frac{3}{2}$.

There is no upper bound: for $a = b = 1$ and $c \to \infty$ the left side tends to $\infty$.

Going deeper

For a student

Nesbitt's inequality is a standard first example of a symmetric inequality in three variables. It has many proofs: by the means inequalities, by the rearrangement inequality, or by clearing denominators and using Muirhead's inequality. Comparing the proofs is a good way to learn the toolbox of elementary inequalities.