## Proof Exercise #2 Continuation 1

proving the Geometric Mean - Arithmetic Mean for the most particular case

Yes.

We will fully exploit the fact that the square of a real number is never negative.

How do you think we will do that, choosing from the list below?

Since neither x nor y is negative, the square root of each is a real number. Let p be the square root of x, and let q be the square root of y. We will reference p and q in the list below.

choice #1: Consider the sum of the cubes of p and q.

choice #2: Consider the quantity (p + q)(x + y).

choice #3: Consider the quantity (p(q +1) + (p + 1)q).

choice #4: Consider the square of the quantity ((p+1)q + p(q + 1)).

choice #5: Consider the square of the quantity (p - q).

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We will fully exploit the fact that the square of a real number is never negative.

How do you think we will do that, choosing from the list below?

Since neither x nor y is negative, the square root of each is a real number. Let p be the square root of x, and let q be the square root of y. We will reference p and q in the list below.

choice #1: Consider the sum of the cubes of p and q.

choice #2: Consider the quantity (p + q)(x + y).

choice #3: Consider the quantity (p(q +1) + (p + 1)q).

choice #4: Consider the square of the quantity ((p+1)q + p(q + 1)).

choice #5: Consider the square of the quantity (p - q).

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