Hello guys, how you doing? This time let’s talk about Cauchy-Schwarz inequality, it’s a well known inequality and useful across multiple branches of mathematics. Actually, it’s a special case of Hölder’s inequality. In math contest, it’s really useful, so let’s start with the definition.

## Definition

Let $a_{i}, b_{i}$, for $i = 1, 2, ..., n$ be real numbers. Then:

Equality occurs if and only if there exists $c \in \mathbb{R}$ such that $b_{i} = ca_{i}$, for $i=1, 2, ..., n$

## An easy proof for Cauchy-Schwarz Inequality

There a lot of proofs about it, I think that below is the easiest way.

Let:

Being a sum of squares, $F(x)$ is always non-negative. Now we expand it and collect terms:

The discriminant is $\leq 0$, computing the discriminant we have:

Dividing by 4 and rearranging yields Cauchy-Schwarz Inequality. It holds when $F$ has a real root (repeated of course). From the first form of $F$ and using the fact that the sum of squares equal to 0 only when each square equals to 0, we have $a_{i}x - b_{i} = 0 \Rightarrow a_{i}/b_{i} = x$ for all $1 \leq i \leq n$ that is when $a_{i}/b_{i}$ is constant for all $i$.

## A Trigonometric Proof

For the graph we know the following:

$\sin(a) = \frac{y_{1}}{\sqrt{x_{1}^2 + y_{1}^2}}$, $\hspace{1.3cm}$ $\cos(a) = \frac{x_{1}}{\sqrt{x_{1}^2 + y_{1}^2}}$

$\sin(b) = \frac{y_{2}}{\sqrt{x_{2}^2 + y_{2}^2}}$, $\hspace{1.3cm}$ $\cos(b) = \frac{x_{2}}{\sqrt{x_{2}^2 + y_{2}^2}}$

Applying trigonometric identities:

$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$

Then:

We know also that $\sin^2(a+b) + \cos^2(a+b) = 1$

So:

Therefore

$(x_{1}y_{2} + x_{2}y_{1})^2 \leq (x_{1}^2 + y_{1}^2)(x_{2}^2 + y_{2}^2),$ which is the two-variable Cauchy-Schwarz Inequality $\ \ \ \mathbb{Q.E.D}$

## Solving problems

Lets begin with an easy problem.

$1.-$ Show that for $x, y, z > 0$, we have

Let’s make

Then, for Cauchy-Schwarz Inequality

$2.-$ (APMO’ 1991) For positive reals $\{ a_{i}, b_{i}\}$ such that $a_{1} + a_{2} + \cdots + a_{n} = b_{1} + b_{2} + \cdots + b_{n}$, show that

We assume as true the last sentence and make an artifice.

We know that $a_{1} + \cdots + a_{n} = b_{1} + \cdots + b_{n}$, then instead of $2(a_{1} + a_{2} + \cdots + a_{n})$ we write $a_{1} + \cdots + a_{n} + b_{1} + \cdots + b_{n}$ Also, let’s make $a_{i} + b_{i} = c_{i}$ for $i = 1, \cdots, n$

Let’s say that before the last sentence, there was a replacement of $a_{i} \rightarrow \frac{a_{i}}{\sqrt{c_{i}}}$ and $c_{i} \rightarrow \sqrt{c_{i}}$ for $1 \leq i \leq n$, then the inequality could have been read as:

We notice that the last sentence is in fact the Cauchy-Schwarz Inequality, then we can conclude that:

But there is something else, we notice that

is a consequence of Cauchy-Schwarz Inequality, this lemma is known as Titu’s Lemma. Which is a special form helps tackle a lot of optimization problems involving squares in the numerator, immediately.

$3.-$ (China Mathematical Olympiad 2004) For a given integer $n \geq 2$, suppose positive integers $a_{i}(i=1, 2, \cdots, n)$ satisfy $a_{1} \leq \cdots \leq a_{n}$ and $\sum_{i=1}^{n}\frac{1}{a_{i}} \leq 1$. Prove that, for any real number $x$, the following inequality holds,

For $x^2 \geq a_{1}(a_{1}-1)$, from $\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 1$ we have

For $% $, using Cauchy-Schwarz Inequality, we have

Futher, for positive integers $% $, we have $a_{i+1} \geq a_{i}+1$ and

For $i=1, \cdots, n-1$. So