Inequality between means: show that 3a^4 - 4a^3b + b^4 ≥ 0, for all real numbers a, b.
- Thread starter Manf
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What are your thoughts? What have you tried? ("Read Before Posting") What sort of math class generated this homework exercise? What topics has the class recently studied?
Please be complete, so we can see where you're getting stuck. Thank you!
Eliz.
I studied inequality between the means, I tried to do AM-GM and AM-HM, between the terms 3a^4 and 4a^3b, because I added b^4 to both sides of the inequality, getting 3a^4 + 4a^3b greater or equal to b^4 and I tried to make AM greater than or equal to GM and AM greater than or equal to HM, but to no avail, no sentences resulting in 3a^4 + 4a^3b.
Now that you stated in post #5 that you managed to do it, consider this alternate method:
\(\displaystyle 3a^4 - 4a^3b + b^4 \ \ vs. \ \ 0\)
\(\displaystyle 2a^4 + a^4 - 4a^3b + b^4 - 2a^2b^2 + 2a^2b^2 \ \ vs. \ \ 0\)
\(\displaystyle (a^4 - 2a^2b^2 + b^4) + (2a^4 - 4a^3b + 2a^2b^2) \ \ vs. \ \ 0\)
\(\displaystyle (a^2 - b^2)^2 + 2a^2(a^2 - 2ab + b^2) \ \ vs. \ \ 0\)
\(\displaystyle (a^2 - b^2)^2 + 2a^2(a - b)^2 \ \ vs. \ \ \ge \ 0 \)