Find and correct the error in the following proof. Claim: for all positive x, y

Question

Find and correct the error in the following proof. Claim: for all positive x, y elementof R, the arithmetic mean x + y/2 is greater than or equal to the geometric mean squareroot xy. "Proof": Consider the real number x - y. Certainly its square is nonnegative: (x - y)^2 greaterthanorequalto 0. Expanding the square, we get x^2 - 2xy + y^2 greaterthanorequalto 0 DoubleRightArrow x^2 + y^2 greaterthanorequalto 2xy. Then take the square root of both sides to get x + y greaterthanorequalto 2 squareroot xy, and divide by 2 to get the desired inequality we wish to show x + y/2 greaterthanorequalto squareroot xy.

Explanation / Answer

Given proof is: Consider the real number x-y. Certainly it's square is non-negative: (x-y)2 >=0. Expanding the square, we get x2-2xy+y2 >= 0 => x2+y2 >= 2xy. Then we take the square root of both sides to get x+y>=2xy, and divide by 2 to get the deisred inequality we wish to show: (x+y)/2 >= (xy)

The error is in the bold marked line of the proof.

The square root of x2+y2 is not x+y, nor is the square root of 2xy, 2.xy

So we add 2xy to x2+y2 to get (x+y)2 and to 2xy to get 4xy

=> (x+y)2 >= 4xy

Now taking square root,

x+y >= 2xy

Dividing by 2 we get the result. So the proof is corrected as:

Consider the real number x-y. Certainly it's square is non-negative: (x-y)2 >=0. Expanding the square, we get x2-2xy+y2 >= 0 => x2+y2 >= 2xy. Adding 2xy both sides, we get x2+y2+2xy >= 2xy+2xy => (x+y)2 >= 4xy. Then we take the square root of both sides to get x+y>=2xy, and divide by 2 to get the deisred inequality we wish to show: (x+y)/2 >= (xy).

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