By using AM-GM twice and multiplying the results, we can easily show that

If $a+b+c=1$ then $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9 \tag 1$$

Now the method below also seems to be valid in each step, yet I cannot see the reason why this proves a different inequality!

$$a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c} \geq 6\tag 2$$ since $x+\frac{1}{x} \geq 2$ for all $x>0$.

So $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 5\tag 3$$

I suspect it may have to do with the possibility of $a$, $b$ or $c$ possibly being negative, hence the method doesn't work since $x+\frac{1}{x} \geq 2$ only if $x>0$, but what if they were all positive quantities?

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    $\begingroup$ I have another inequality, as $a,b,c\leq 1$ then $\frac 1a , \frac 1 b, \frac 1 c\geq 1$ hence $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3\tag 4 $$ $\endgroup$
    – Elaqqad
    Aug 5 '15 at 15:23
  • $\begingroup$ I'm not completely sure about the idea of both inequalities holding. For example, according to my third equation, the quantity $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ could be 6 for example, as it is greater than 5. However, this contradicts $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9$. This is my main issue with having two inequalities holding. My understanding is also that equality occurs at the minimum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ so wouldn't having two inequalities imply two minimal values? $\endgroup$
    – Trogdor
    Aug 5 '15 at 15:30
  • $\begingroup$ @Trogdor, the main idea is that the inequalities $(1)$ and $(3)$ are both correct (because you proved them ) . This means that the LHS , the quantity $\frac 1a+\frac 1b+\frac1c$ is always greater than both $5$ and $9$. So for example the quantity would never be $6$. or any number between $6$ and $9$. $\endgroup$
    – Elaqqad
    Aug 5 '15 at 15:41
  • $\begingroup$ If I proved that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 5$, then there must exist some $(a,b,c)$ such that equality occurs. In this case there is no such triplet, so surely there must be something wrong with the proof somewhere? A perfectly correct proof can not lead to an incorrect implication, can it? Another question I had about your response is "How would we know what the 'real' lower bound is?" Say I approached finding the minimum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ by using the method in equation 2. How would I know this yields a 'false' lower bound? $\endgroup$
    – Trogdor
    Aug 5 '15 at 15:47
  • $\begingroup$ Note that if an inequality is true, it does not necessarily imply equality must hold. For eg $2\ge1, x^2+y^2+1\ge 0$ are all perfectly valid. When an inequality is true and equality is also possible for some case, then you have found a best bound - min or max of the expression. $\endgroup$
    – Macavity
    Aug 5 '15 at 16:21

You proved that $\frac1a+\frac1b+\frac1c \geq 5$, which is consistent with $\frac1a+\frac1b+\frac1c \geq 9$. You just didn't use an inequality that was strong enough. I could also say that $\frac1a > 0$, $\frac1b > 0$ , $\frac1c > 0$, since they are all positive. Thus proving $\frac1a+\frac1b+\frac1c > 0$.

If you use $x+\frac{1}{x}\geq2$, we have equality only if $x=1$. However we can't have $a=b=c=1$ because of the constraint $a+b+c=1$. Therefore the method doesn't work.

To actually solve it you can use the solution below. I've put it in a spoiler box if you didn't want to see it yet. Also this assumes $a,b,c>0$ but that is required.

Use AM-HM: $$\frac13 = \frac{a+b+c}{3} > \frac{3}{\frac1a+\frac1b+\frac1c}$$ Therefore $$\frac19 > \frac{1}{\frac1a+\frac1b+\frac1c}$$ Therefore $$ \frac1a+\frac1b+\frac1c > 9$$

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    $\begingroup$ And it's the best lower bound : take $a=b=c = \frac{1}{3}$ to achieve it $\endgroup$
    – Tryss
    Aug 5 '15 at 15:24
  • $\begingroup$ I don't see what do you mean by "To actually solve it", and the OP's question is very interesting , sometimes we all get inequalities which are not sharp enough , so I think you need to add more explanation then just providing another proof. (the op has already one). I don't understand either what do you mean by "You didn't use an inequality that was strong enough" $\endgroup$
    – Elaqqad
    Aug 5 '15 at 15:26
  • $\begingroup$ @Elaqqad I have provided the explaination in the first paragraph. $\endgroup$
    – wythagoras
    Aug 5 '15 at 15:28

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