How can I prove the inequality A(B-A) <= (B/2)^2 for 0 <= A <= B?

[ausdreamer]

Homework Statement

If 0 <= A <= B, prove that: A(B-A) <= (B/2)^2

Homework Equations

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The Attempt at a Solution

I've been blindly rearranging the terms trying to see a way to prove this but due to my complete lack of experience in proofs, I'm hoping someone here can give a little push in a helpful direction.

 

[babita]

try opening the brackets and taking all the terms to one side. it'll become square of a number.
but i dnt understand how 0>=a>=b are essential conditions for this. square of any real no would always be positive

 

[babita]

sry i typed the inequality wrong

 

[tms]

The cases where $A = 0$ and $A = B$ should be obvious. For the rest, $0 < A < B$, think geometrically.

 

[asteriatic]

As a scientist, the first step in proving this inequality is to define the variables. In this case, A and B represent any real numbers, with the added constraint that A must be less than or equal to B. This means that A and B can be positive, negative, or zero, but A must always be less than or equal to B. Now, we can rewrite the inequality as:

A(B-A) ≤ (B/2)^2

Next, we can expand the right side of the inequality by squaring the expression:

A(B-A) ≤ B^2/4

Now, let's consider the left side of the inequality. We can use the distributive property to expand the expression:

A(B-A) = AB - A^2

Since A is less than or equal to B, we know that A^2 is less than or equal to AB. This means that AB - A^2 will always be positive or zero. Therefore, we can rewrite the left side of the inequality as:

A(B-A) ≤ AB

Now, we can substitute this into our original inequality to get:

AB ≤ B^2/4

Since we know that A is less than or equal to B, we can divide both sides of the inequality by B to get:

A ≤ B/4

Finally, since we know that A is greater than or equal to 0 (given in the problem statement), we can divide both sides by A to get:

1 ≤ B/4A

Since A is greater than or equal to 0, 4A must also be greater than or equal to 0. Therefore, we can multiply both sides by 4A without changing the direction of the inequality:

4A ≤ B

And since A is less than or equal to B, we know that this inequality is always true. Therefore, we have proven that A(B-A) ≤ (B/2)^2 for 0 ≤ A ≤ B.