Consider the quadratic in t,
(tx1 + y1)² + (tx2 + y2)² + (tx3 + y3)²
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.
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)
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
The Quadratic Formula
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