2
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $$ \frac1a+\frac1b+\frac1c > 9$$

$\endgroup$
3
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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