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In the course of working on a problem on normal numbers, I ran into the following inequality:

Let $n$ be a positive integer, $p_i$, $1\le i \le n$, be real numbers, and $q_i$, $1\le i \le n$, be positive numbers. Then, $$ \frac{\left( \sum_{i=1}^n p_i \right)^2}{\sum_{i=1}^n q_i} \le \sum_{i=1}^n \frac{p_i^2}{q_i}$$ with equality if and only if all the fractions $p_i/q_i$ are equal.

I believe I have found a proof of this using a moderately straight-forward induction on $n$; however, after having worked with this inequality for a while, I couldn't shake the feeling that either (1) someone else has discovered this inequality before; or, (2) that there should be an even more direct derivation of this result from a well-known inequality. Has anyone seen an inequality like this before? Is there a quick derivation of the inequality that I missed?

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migrated from Aug 15 '13 at 9:48

This question came from our site for professional mathematicians.

It goes by the name Titu's lemma as far as I know – Ákos Somogyi Aug 25 '15 at 17:19

Cauchy-Schwarz with $p_i/\sqrt{q_i}$ and $\sqrt{q_i}$.

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Nice. That's much slicker than my answer... – Steve Kass Aug 15 '13 at 5:20

This is Jensen's inequality in disguise, for the concave-up function $f(x)=x^2$.

Let $x_i=\frac{p_i}{q_i}$, $Q={\sum{q_i}}$ and $w_i=\frac{q_i}Q$. Note that $\sum{w_i}=1$.

Then rewrite the left and right-hand sides of your inequality: $$\frac{\left( \sum_{i=1}^n p_i \right)^2}{\sum_{i=1}^n q_i} = \frac{\left( \sum_{i=1}^n x_iq_i \right)^2}{Q}= \frac{\left( \sum_{i=1}^n x_iw_iQ \right)^2}{Q}= Q\left( \sum_{i=1}^n x_iw_i\right)^2$$

$$\sum_{i=1}^n \frac{p_i^2}{q_i}=\sum_{i=1}^n \frac{(x_iq_i)^2}{q_i}=\sum_{i=1}^n (x_i)^2q_i=Q\sum_{i=1}^n (x_i)^2w_i$$

Your inequality is now Jensen's inequality: $$\left( \sum_{i=1}^n x_iw_i\right)^2 \le \sum_{i=1}^n (x_i)^2w_i$$

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