MHB Prove: $ab+ 2a^2b^2 \le a^2 + b^2 + ab^3$ for 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1

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    2016
AI Thread Summary
The discussion centers on proving the inequality $ab + 2a^2b^2 \le a^2 + b^2 + ab^3$ for the variables a and b constrained between 0 and 1. Participants are encouraged to submit their proofs in accordance with the Problem of the Week guidelines. Two members, kaliprasad and lfdahl, successfully provided correct solutions to the problem. Kaliprasad's approach is highlighted, though the details of the solution are not included in the summary. The thread emphasizes collaborative problem-solving within the mathematical community.
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Here is this week's POTW:

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Prove that $ab+ 2a^2b^2\le a^2 +b^2 +ab^3$ for all reals $0\le a \le 1$ and $0\le b\le 1$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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anemone has asked me to stand in for her this week.

Congratulations to the following members for correctly answering the given problem:

  • kaliprasad
  • lfdahl

kaliprasad's solution is as follows:

Because $a <=1$ and b is positive hence $2ab<=2b\cdots(1)$
Now $a^2+b^2+ab^3-ab-2a^2b^2$
$= a^2+b^2 - 2ab + ab^3 +ab - 2a^2b^2$
$= (a-b)^2 + ab(b^2 + 1 - 2ab)$
$>=(a-b)^2 + ab(b^2 +1 - 2b$ using (1)
$>= (a-b)^2 + ab(1-b)^2$
$>=0$
or $a^2+b^2 + ab^3 >= ab + 2a^2b^2$
or $ab + 2a^2b^2<= a^2+b^2 + ab^3$
 
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