Lesson AM-GM inequality
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It is a fact that for any 2 nonnegative numbers, the arithmetic mean is always greater than or equal to the geometric mean of the 2 numbers. <H2>Proof</H2> The arithmetic mean of 2 numbers x and y is {{{(x+y)/2}}} and the geometric mean is {{{sqrt(xy)}}} So we are trying to prove that {{{(x+y)/2>=sqrt(xy)}}} Square both sides to get {{{(x^2+2xy+y^2)/4>=xy}}} Move all terms to one side. {{{((x^2+2xy+y^2)/4)-xy>=0}}} Find common denominator. {{{((x^2+2xy+y^2)/4)-4xy/4>=0}}} {{{(x^2-2xy+y^2)/4>=0}}} Multiply both sides by 4. {{{x^2-2xy+y^2>=0}}} Factor left side. {{{(x-y)^2>=0}}} Since the square of any real number is always greater than or equal to 0, the proof is done.
