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Exercise 1020Algebra › InequalitiesL1Added 3 Nov 2026

Titu's lemma (Cauchy in Engel form)

#Titu#Engel form#Cauchy-Schwarz

Statement

Prove Titu's lemma: for xi∈Rx_i\in\mathbb{R} and yi>0y_i>0,
∑i=1nxi2yi≥(∑xi)2∑yi.\sum_{i=1}^n\frac{x_i^2}{y_i}\ge\frac{\left(\sum x_i\right)^2}{\sum y_i}.

Hint

Apply Cauchy-Schwarz to the vectors (yi)(\sqrt{y_i}) and (xi/yi)(x_i/\sqrt{y_i}).

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Solution

Cauchy-Schwarz. Apply Cauchy-Schwarz to (y1,…,yn)(\sqrt{y_1},\ldots,\sqrt{y_n}) and (x1/y1,…,xn/yn)(x_1/\sqrt{y_1},\ldots,x_n/\sqrt{y_n}):
(∑iyi⋅xiyi)2≤(∑iyi)(∑ixi2yi).\left(\sum_i \sqrt{y_i}\cdot\frac{x_i}{\sqrt{y_i}}\right)^2\le\left(\sum_i y_i\right)\left(\sum_i\frac{x_i^2}{y_i}\right).
The left side is (∑xi)2(\sum x_i)^2.

Rearranging.
(∑xi)2∑yi≤∑ixi2yi.\frac{(\sum x_i)^2}{\sum y_i}\le\sum_i\frac{x_i^2}{y_i}.

Conclusion. Titu's lemma holds, with equality iff the two vectors are proportional, i.e. xi/yix_i/y_i is constant.