## The Cauchy-Schwarz Inequality

### (x_{1}² + x_{2}² + x_{3}²)(y_{1}² + y_{2}²
+ y_{3}²) ≥ (x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3})²,
for reals x_{i} and y_{j}

Proof:

Consider the quadratic in t,

(tx_{1} + y_{1})² + (tx_{2} + y_{2})² + (tx_{3}
+ y_{3})²

If you expand it out, it becomes

t²(x_{1}²+x_{2}²+x_{3}²) + 2t(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})
+ (y_{1}²+y_{2}²+y_{3}²)

Now, we know this can't possibly be negative, because it's a sum of three
squares and squares are always positive.

If we have a quadratic of the form at² + bt + c, then this is zero when t = (
-b � sqrt(b² - 4ac) )/(2a). This is the
quadratic formula.

If the quadratic in t, above, had two distinct roots, then it would be
negative between the roots (since a=(x_{1}²+x_{2}²+x_{3}²)
is nonnegative). But we know this can't be the case, because it's a sum of
squares. So the quadratic formula must give either only one root, which
can only happen if b² - 4ac = 0, or no real roots at all, which happens if b² -
4ac < 0.

So, let's calculate this quantity, b²-4ac, which is called the *
discriminant*
of the quadratic:

4(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})²
- 4(x_{1}²+x_{2}²+x_{3}²)(y_{1}²+y_{2}²+y_{3}²)

This discriminant must be less than or equal to zero, *i.e.*

(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})²
≤ (x_{1}²+x_{2}²+x_{3}²)(y_{1}²+y_{2}²+y_{3}²)

which is the Cauchy-Schwarz inequality.

### Uses of the Cauchy-Schwarz Inequality

**If there are 3 positive numbers, x,y,z such that x+y+z ≤ 3,**

prove that 1/x + 1/y +1/z ≥ 3

Let x_{1}=sqrt(x), y_{1}=sqrt(1/x), x_{2}=sqrt(y), y_{2}=sqrt(1/y),
x_{3}=sqrt(z), y_{3}=sqrt(1/z).

Then (x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})²
= (1+1+1)² = 9.

From the Cauchy-Schwarz Inequality, you have

9 ≤ (x+y+z)(1/x+1/y+1/z)

Now divide both sides by x+y+z (which is legal because you know this is a
positive number). You know 9/(x+y+z) is at least 3, because the
denominator is at most 3, so

3 ≤ 9/(x+y+z) ≤ (1/x+1/y+1/z)

### Related pages in this website

The AM-GM
Inequality: the Arithmetic Mean of positive numbers is always greater than
the Geometric Mean. This is proved using Jensen's Inequality.

The Triangle Inequality

The Chebyshev Sum Inequality

Puzzles

The Quadratic Formula

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Graeme McRae.