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

[anemone]

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!

 

[MarkFL]

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:

Spoiler

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 $ab + 2a^2b^2<= a^2+b^2 + ab^3$