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

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To prove the inequality A(B-A) <= (B/2)^2 for 0 <= A <= B, one approach is to rearrange the terms and analyze the expression geometrically. The conditions 0 <= A <= B are crucial as they define the boundaries for A and B, ensuring that the product A(B-A) remains non-negative. When A equals 0 or B, the inequality holds trivially. For the case where 0 < A < B, a geometric interpretation can help visualize the relationship between the variables. Understanding these aspects is key to successfully proving the inequality.
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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.
 
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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
 
sry i typed the inequality wrong
 
The cases where A = 0 and A = B should be obvious. For the rest, 0 &lt; A &lt; B, think geometrically.
 
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