Prove that the inequality
(a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2
holds for all real numbers a and b.
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!