TITLE: How to prove an Inequality
QUESTION [10 upvotes]: I'm beginner with proofs and I got the follow exercise:

Prove the inequality $$(a + b)\Bigl(\frac{1}{a} + \frac{4}{b}\Bigr) \ge 9$$ when $a > 0$ and $b > 0$. Determine when the equality occurs.

I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ? 
Thanks in advance.

REPLY [3 votes]: Another one...
$ (a+b)({1 \over b} + {4 \over a}) \ge 9 \qquad \qquad  $
This requires that also $a,b\ne 0$  and thus $a,b \gt 0$ as from the problem-definition    
Then multiplication with ab does not introduce inconsistencies and we can do
$ (a+b)({1 \over a} + {4 \over b}) \ge 9 \qquad \qquad // *ab $
$ (a+b)(b + 4a) \ge 9ab \qquad \qquad   // $  expand
$ 4a^2+5ab+b^2  \ge 9ab \qquad \qquad   // - 9ab $
$ 4a^2-4ab+b^2  \ge 0 \qquad \qquad    $      
Finally:
(1) $ \qquad (2a-b)^2  \ge 0 \qquad \qquad    $   This is always true     
(2) $ \qquad $ The case of equality is now obvious...