Find the minimum value of $\frac{1}{a} + \frac{9}{b} + \frac{36}{c}$ given that $a+b+c=2$.

Question

Find the minimum value of $\frac{1}{a} + \frac{9}{b} + \frac{36}{c}$ given that $a+b+c=2$ and $a,b,c > 0$.

So this looks like an $AM \ge HM$ problem but I'm not able to get the correct answer.

By weighted $HM$ concept:

$\frac{a+b+c}{n} \ge \frac{n}{\frac{1}{a} + \frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b} +\frac{1}{c} +\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}}$

basically:

$$\frac{a+b+c}{n} \ge \frac{n}{\frac{1}{a} + \frac{1}{b}+....+\frac{1}{b} +\frac{1}{c} +\frac{1}{c}+\frac{1}{c}....+\frac{1}{c}}$$

Let $f(x) = \frac{1}{a} + \frac{1}{b}+....+\frac{1}{b} +\frac{1}{c} +\frac{1}{c}+\frac{1}{c}....+\frac{1}{c}$ (This is what we want to find.)

where $\frac{1}{b}$ is occurring $9$ times and $\frac{1}{c}$, $36$ times.

so that gives us:

$$\frac{2}{46} = \frac{46}{f(x)}$$

giving us: $f(x) = 23 \times 46$

But this is not the answer. I am not getting what I am doing wrong. Any help would be appreciated.

what happens when $a\to 0$ ? there is no maximum did you mean minimum value ? — Albus Dumbledore, Mar 31 '21 at 13:10
https://math.stackexchange.com/questions/3619012/how-prove-this-inequality-fraca2b-fracb2c-fracc2a6/3636210#3636210 — Aderinsola Joshua, Mar 31 '21 at 13:22
It must be the minimum value as $AM ≥ HM$ can only find the minimum, not the maximum. If the OP has intended differently, they can edit the post themselves. — Toby Mak, Mar 31 '21 at 13:29
There are two deleted answers about the minimum (only users with >10k rep can see them). — Toby Mak, Mar 31 '21 at 13:30
Yes, I meant minimum value. Sorry for the mistake, I'll make sure I don't commit such errors again. — , Mar 31 '21 at 13:39
@AlbusDumbledore You can undelete your post now. Cauchy Schwarz on $(1/a + 9/b + 36/c)(a + b + c) ≥ (1 + 3 + 6)^2$ works, as Albus has said. — Toby Mak, Mar 31 '21 at 13:41

score 2 · Accepted Answer · answered Mar 31 '21 at 13:06

2

Hint: Use Cauchy schwarz $$(\frac{1}{a} + \frac{9}{b} + \frac{36}{c})(a+b+c)\ge {(1+3+6)}^2$$

answered Mar 31 '21 at 13:06

Albus Dumbledore

11,860

That will give us the minimum value. – Martund Mar 31 '21 at 13:08
@Martund the question has been changed in support of my answer now :) – Albus Dumbledore Mar 31 '21 at 13:50
That's good, Albus. – Martund Mar 31 '21 at 14:44

Find the minimum value of $\frac{1}{a} + \frac{9}{b} + \frac{36}{c}$ given that $a+b+c=2$.

1 Answers1