# Why does this method 'prove' a different inequality?

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?

• I have another inequality, as $a,b,c\leq 1$ then $\frac 1a , \frac 1 b, \frac 1 c\geq 1$ hence $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3\tag 4$$ Aug 5 '15 at 15:23
• 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? Aug 5 '15 at 15:30
• @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 $9$. Aug 5 '15 at 15:41
• 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? Aug 5 '15 at 15:47
• 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. 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 $$\frac1a+\frac1b+\frac1c > 9$$
• And it's the best lower bound : take $a=b=c = \frac{1}{3}$ to achieve it Aug 5 '15 at 15:24