4
$\begingroup$

I've been trying to find the minimum value of $\frac{x^2}{x-9}$ when x>9 using AM-GM inequality but am unable to do so. The problem is trivial using calculus but I would like to see it done using AM-GM. I am aware that the answer is $36$.

$\endgroup$
2
  • 1
    $\begingroup$ Check your equation. The left-hand limit of the linked function is $-\infty$ while the right-hand limit is $\infty$. There is no minimum value. Perhaps you mean on a specific interval? $\endgroup$
    – JMoravitz
    Jan 4 '16 at 3:29
  • 1
    $\begingroup$ Ah, sorry about that. I've edited the question to only consider when x>9 $\endgroup$
    – Jason G.
    Jan 4 '16 at 3:36
8
$\begingroup$

Let $t=x-9$, then $$ \frac{x^2}{x-9}=\frac{(t+9)^2}{t} = t+\frac{81}{t}+18 \ge 2 \sqrt{81}+18=36 $$

$\endgroup$
5
  • 2
    $\begingroup$ Interesting. I would ask how the interval given is taken into account but, by making the substitution t = x-9 and taking note that x>9, this ensures that t>0. Because t>0, AM-GM inequality holds (only applies to non-negative reals). Very nice--thank you! $\endgroup$
    – Jason G.
    Jan 4 '16 at 3:54
  • $\begingroup$ On second thought, I wonder why doing t + 81/t + 18 >= 3(81*18)^(1/3) produces a different minimum value? 36 is the answer but trying to do it in 3 variables produces a result of about 34, which is wrong. What is wrong with approaching it like so? Of course, the inequality holds, but doing it in three-variables does not provide the actual minimum of the function in x>9. $\endgroup$
    – Jason G.
    Jan 4 '16 at 14:38
  • $\begingroup$ @JasonG. Take note that the equality holds if and only if $t=\frac{81}{t}=18$, which is impossible. $\endgroup$
    – wangjiezhe
    Jan 5 '16 at 1:20
  • $\begingroup$ I am aware that the equality case does not hold, but what does that have to do with the three-variable case? Sorry, I'm missing something. $\endgroup$
    – Jason G.
    Jan 5 '16 at 3:32
  • $\begingroup$ It means that $\ge 34$ is always true, but you cannot get to this point, so it makes no sense to this question. $\endgroup$
    – wangjiezhe
    Jan 5 '16 at 9:17

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.