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 Math Help > Inequalities > Inequality Methods > Cauchy-Schwartz

## The Cauchy-Schwarz Inequality

### (x1² + x2² + x3²)(y1² + y2² + y3²) ≥ (x1y1 + x2y2 + x3y3)², for reals xi and yj

Proof:

(tx1 + y1)² + (tx2 + y2)² + (tx3 + y3

If you expand it out, it becomes

t²(x1²+x2²+x3²) + 2t(x1y1+x2y2+x3y3) + (y1²+y2²+y3²)

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=(x1²+x2²+x3²) 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(x1y1+x2y2+x3y3)² - 4(x1²+x2²+x3²)(y1²+y2²+y3²)

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

(x1y1+x2y2+x3y3)² ≤ (x1²+x2²+x3²)(y1²+y2²+y3²)

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 x1=sqrt(x), y1=sqrt(1/x), x2=sqrt(y), y2=sqrt(1/y), x3=sqrt(z), y3=sqrt(1/z).

Then (x1y1+x2y2+x3y3)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 webmaster and author of this Math Help site is Graeme McRae.