Proof using Jensen's Inequality:
Jensen's Inequality is
∑(f(xi))/n ≤ f(∑(xi)/n) if f is concave (i.e. f''(x)<0, which is sometimes called "concave down"),
∑(f(xi))/n ≥ f(∑(xi)/n) if f is convex (i.e. f''(x)>0, which is sometimes called "concave up"),
with equality iff x1 = x2 = ... = xn.
ln(x) is concave, so by Jensen's Inequality,
(ln(a1) + ln(a2) + ... + ln(an))/n ≤ ln((a1+a2+...+an)/n),
Taking the exponential function of both sides,
geometric mean(a1,a2,...,an) ≤ arithmetic mean(a1,a2,...,an)
Given n positive reals a1,a2,...,an, and n
weights w1,w2,...,wn such that 0≤wi≤1
w1a1+w2a2+...+wnan ≥ a1w1 a2w2 ... anwn
Proof Outline: same as above, using the weighted version of Jensen's Inequality; or use the ordinary AM-GM inequality for rational weights (put all the weights over a common denominator, d, and then show that the AM of the d numbers (many identical) is never less than the GM of those same numbers); and then extend to irrational weights by continuity.
Mathworld -- Jensen's Inequality
The Cauchy-Schwarz inequality
The Triangle Inequality
The Chebyshev Sum Inequality
The Quadratic Formula
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