Prove that the inequality

(a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2

holds for all real numbers a and b.

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- Sep 1st 2005, 03:53 AMardnasProving Inequalities
Prove that the inequality

(a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2

holds for all real numbers a and b. - Sep 1st 2005, 02:25 PMrgep
OK, let's imagine that we're working on a piece of scrap paper, not writing out the finished solution. For inequalities you have to be careful when trying to reverse the steps.

(a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2

<=>

4a^2 + 29ab + 7b^2 >= 3a^2 + 33ab + 2b^2

<=>

a^2 - 4ab + 5b^2 >= 0

<=>

(a-2b)^2 + b^2 >= 0

Now this last line is true, and each step in the chain is reversible. So we take our answer paper and write down

(a-2b)^2 + b^2 >= 0

expand

hence a^2 - 4ab + 5b^2 >= 0

add 3a^2 + 33ab + 2b^2 to both sides

hence

4a^2 + 29ab + 7b^2 >= 3a^2 + 33ab + 2b^2

factorise LHS

hence (a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2

Throw away the scrap paper and pretend that you thought of it all in that order!