Prove that if x and y are nonnegative real numbers, then (x+y)/2 ≥ sqrt(xy).
Use the technique of working backward from the desired conclusion. Where in the proof do we use the fact that x and y are nonnegative?
Well, this is one of those fancy math proofs that looks totally scary at first but can actually be broken down into simpler steps. Let's tackle it together, shall we?
To prove that (x+y)/2 ≥ sqrt(xy), we'll work our way backward from the desired conclusion. So, let's start from the end and see how far we can go:
Desired conclusion: (x+y)/2 ≥ sqrt(xy)
Now, let's square both sides of the inequality to make our lives a bit easier:
((x+y)/2)^2 ≥ (sqrt(xy))^2
Expanding the left side:
(x^2 + 2xy + y^2)/4 ≥ xy
Now, multiplying both sides by 4 to get rid of that pesky denominator:
x^2 + 2xy + y^2 ≥ 4xy
Simplifying the left side:
x^2 - 2xy + y^2 ≥ 0
And now we're getting somewhere! This is the point where we use the fact that x and y are nonnegative. Since x and y are both nonnegative, their square terms (x^2 and y^2) are also nonnegative.
So, we have:
x^2 - 2xy + y^2 ≥ 0
Great! We've proved it! The fact that x and y are nonnegative is used here to guarantee that the left side of the inequality is nonnegative.
Voila! We've successfully proved that (x+y)/2 ≥ sqrt(xy) using the technique of working backward from the desired conclusion. And hey, isn't math just a wonderfully mysterious land of numbers and symbols?
To prove the inequality (x+y)/2 ≥ √(xy) for nonnegative real numbers x and y, we will use the technique of working backward from the desired conclusion.
Let's start with the desired conclusion:
(x+y)/2 ≥ √(xy)
We square both sides of the inequality to eliminate the square root:
[(x+y)/2]^2 ≥ (xy)
Expanding the left side of the inequality:
(x^2 + 2xy + y^2)/4 ≥ (xy)
Multiplying both sides by 4 to eliminate the fraction:
x^2 + 2xy + y^2 ≥ 4xy
Rearranging the terms:
x^2 + y^2 - 2xy ≥ 0
Now we need to examine where we used the fact that x and y are nonnegative. In this step, we need to use the fact that the square of any real number is nonnegative. Since x^2 and y^2 are squares, they are always nonnegative.
Therefore, the inequality x^2 + y^2 - 2xy ≥ 0 holds true for any real numbers x and y, regardless of their signs.
Hence, we have proven that if x and y are nonnegative real numbers, then (x+y)/2 ≥ √(xy).