A nice use of $AM GM HM$ inequality

Question

Let $a+b+c = 20$. Then what is the minimum value of $1/a + 9/b + 36/c$ .In this question minimum value is asked. How to approach it as we can find maximum value by applying $GM $ and $HM$ inequality.

Answer 1

We have, $$ a + b + c = 20 $$ rewriting as - $$ a + (b/3 + ... 3 times) + (c/6 + ... 6 times) = 20 \tag 1$$ now, we have to find - $$ 1/a + (3/b + .. 3 times) + (6/c + ... 6 times) \tag 2$$ Since a,b,c are all positive numbers, we can apply AM > HM. Treat (2)/10 as AM and 10/(1) as HM - $$ \frac{(2)}{10} \ge \frac {10}{(1)} $$ Solve for minimum value of (2). Equality holds.