Cauchy-Schwarz InequalityDate: 11/13/1999 at 23:36:15 From: Nick Krupar Subject: Cauchy-Schwarz Inequality Prove: n n n [sum(a_j)(b_j)]^2 <= (sum[j(a_j)^2])(sum[[(b_j)^2]/j]) j=1 j=1 j=1 I've shown that, where <...> is the inner-product: n [sum(a_j)(b_j)]^2 = <(a_1, ..., a_n),(b_1, ..., b_n)>^2 j=1 n n <= (sum[(a_j)^2])(sum[(b_j)^2]) j=1 j=1 I know that: n n sum[(a_j)^2] <= sum[j(a_j)^2] j=1 j=1 n n sum[(b_j)^2] >= sum[[(b_j)^2]/j] j=1 j=1 Date: 11/14/1999 at 08:17:31 From: Doctor Anthony Subject: Re: Cauchy-Schwarz Inequality The Cauchy-Schwarz inequality is that for real numbers ai, bi then (a1.b1 + a2.b2 + .... + an.bn)^2 <= (a1^2 + a2^2 + ... + an^2)* (b1^2 + b2^2 + ... + bn^2) Proof: For any x we have 0 <= (a1.x+b1)^2 + (a2.x+b2)^2 + ... + (an.x + bn)^2 = (a1^2 + ... + an^2)x^2 + 2(a1.b1 + ... + an.bn)x + (b1^2 + ... + bn^2) = Ax^2 + 2Bx + C since y = Ax^2 + 2Bx + C >= 0 for ALL x it follows that the equation Ax^2 + 2Bx + C = 0 cannot have two distinct roots. And so (2B)^2 <= 4AC (discriminant of the quadratic equation <= 0) so B^2 <= AC so (a1.b1 + ... + an.bn)^2 <= (a1^2 + ... + an^2)(b1^2 + ... + bn^2) which is the required inequality. A further inequality that is often quoted is: (x1 + x2 + x3 + ...+ xn)^2 <= n(x1^2 + x2^2 + ... + xn^2) If we take just 3 values, we require to show (x1 + x2 + x3)^2 <= 3(x1^2 + x2^2 + x3^2) x1^2+x2^2+x3^2+2x1.x2+2x2.x3+2x3.x1 <= 3(x1^2 + x2^2 + x3^2) and so 2x1.x2 + 2x2.x3 + 2x3.x1 <= 2x1^2 + 2x2^2 + 2x3^2 0 <= x1^2 - 2x1.x2 + x^2 + x1^2 - 2x1.x3 + x3^2 + x2^2 - 2x2.x3 + x3^2 0 <= (x1-x2)^2 + (x1-x3)^2 + (x2-x3)^2 which is clearly true since each term on the right is >= 0 The same pattern is true if you take higher values of n. You will always be able to put the inequality into the form shown above. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/