Cauchy-Schwarz And Titu Inequalities – Mattia Giuri's bizarre blog

Today I’m going to show two very nice inequalities to know, which really are useful in the mathematical olympiads.

What They Say And Their Proof

The Cauchy-Schwarz inequality asserts that, for real tuples $(x_1, x_2, ..., x_n), (y_1, y_2, ..., y_n)$ we have:

$(\sum_{i=1}^{n}x_i^2)(\sum_{i=1}^{n}y_i^2) \geq (\sum_{i=1}^{n} x_iy_i)^2$

Here’s the proof:

consider the expression

$(\sum_{i=1}^{n}(x_ia + y_i)^2) \geq 0$

It’s obviously true since each term of the sum is non negative. Hence we expand the sum to get:

$(\sum_{i=1}^{n}x_i^2)a^2 + 2a(\sum_{i=1}^{n}x_iy_i) + (\sum_{i=1}^{n}y_i^2) \geq 0$

Since it’s non negative, it has at most one real root, so its discriminant must be less than or equal to 0, which gives us the desired result:

$(\sum_{i=1}^{n}x_iy_i)^2 - (\sum_{i=1}^{n}x_i^2)(\sum_{i=1}^{n}y_i^2) \leq 0$

This proof gives also the equality case: there must exist a real number a such that

$ax_i + y_i = 0, \forall i$

While Cauchy-Schwarz works for all reals, Titu’s Lemma only works for positive reals, in fact it’s a direct consequence of Cauchy-Schwarz where the tuples are $\frac{x_i}{\sqrt{y_i}}, \sqrt{y_i}$

Substituting these we get its final form:

$\sum_{i=1}^{n}\frac{x_i^2}{y_i} \geq \frac{(\sum_{i=1}^{n}x_i)^2}{\sum_{i=1}^{n}y_i}$

An Example For Cauchy-Schwarz

We have three real numbers such that $3x+4y+7z = 74$ .

Find the minimum of $x^2 + y^2 + z^2$

By Cauchy-Schwarz Inequality we get

$(x^2+y^2+z^2)(9+16+49) \geq (3x+4y+7z)^2 = 74^2$

Hence the minimum is 74.

An Example For Titu

Let $a_1, a_2, ..., a_n$ be positive reals and $a_{n+1} = a_1$

Prove that

$2\sum_{i=1}^{n}\frac{a_i^2}{a_i+a_{i+1}} \geq \sum_{i=1}^{n}a_i$

By Titu’s Lemma on the tuples $x_i, x_i+x_{i+1}$ we get

$2\sum_{i=1}^{n}\frac{a_i^2}{a_i+a_{i+1}} \geq 2\frac{(\sum_{i=1}^{n}a_i)^2}{2\sum_{i=1}^{n}a_i} = \sum_{i=1}^{n}a_i$

And we are done.