Exercise 2162Algebra › InequalitiesOlympiadAdded 3 Nov 2026

Generalized Nesbitt inequality

#Nesbitt#generalization#n variables

Statement

Let $x_1,\ldots,x_n>0$ and $S=x_1+\cdots+x_n$ . Prove that
$\sum_{i=1}^n\frac{x_i}{S-x_i}\ge\frac{n}{n-1}.$

Hint

Use Titu's lemma: $\frac{x_i}{S-x_i}=\frac{x_i^2}{x_i(S-x_i)}$ .

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Solution

Titu's lemma.
$\sum\frac{x_i}{S-x_i}=\sum\frac{x_i^2}{x_i(S-x_i)}\ge\frac{(\sum x_i)^2}{\sum x_i(S-x_i)}=\frac{S^2}{S\sum x_i-\sum x_i^2}=\frac{S^2}{S^2-\sum x_i^2}.$

Bound. By Cauchy-Schwarz, $\sum x_i^2\ge S^2/n$ . So
$\frac{S^2}{S^2-\sum x_i^2}\ge\frac{S^2}{S^2-S^2/n}=\frac{n}{n-1}.$

Conclusion. The inequality holds, with equality iff all $x_i$ are equal.

Generalized Nesbitt inequality

Statement

Solution

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