Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the minimum value of the quantity where a , b , c are real positive numbers.

$$\left(\frac{a^2 + 3a + 1}{a}\right) \left(\frac{b^2 +3b + 1}{b}\right)\left(\frac{c^2 + 3c + 1}{c}\right) $$

I think the to get the answer we need to use

A.M.  >= G.M 

How i can achieve this?

share|cite|improve this question
up vote 2 down vote accepted

You can use the fact that $x + \frac{1}{x} \ge 2$, which can be proved using $\text{AM} \ge \text{GM}$, or just completing the square.

share|cite|improve this answer
Thanks. Got it. – vikiiii Mar 6 '12 at 7:19
@vikiiii: You are welcome. – Aryabhata Mar 6 '12 at 7:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.