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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 $36$.

Jason G.
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    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? – JMoravitz Jan 04 '16 at 03:29
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    Ah, sorry about that. I've edited the question to only consider when x>9 – Jason G. Jan 04 '16 at 03:36

2 Answers2

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

wangjiezhe
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    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! – Jason G. Jan 04 '16 at 03:54
  • 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. – Jason G. Jan 04 '16 at 14:38
  • @JasonG. Take note that the equality holds if and only if $t=\frac{81}{t}=18$, which is impossible. – wangjiezhe Jan 05 '16 at 01:20
  • 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. – Jason G. Jan 05 '16 at 03:32
  • It means that $\ge 34$ is always true, but you cannot get to this point, so it makes no sense to this question. – wangjiezhe Jan 05 '16 at 09:17
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wangjiezhe gave a pretty nice and simple solution!

I want to share one more alternate method other than A.M.$\geq G,M.$

For any $x>9,$ we have $$ \frac{x^{2}}{x-9}-36=\frac{x^{2}-36 x+324}{x-9}=\frac{(x-18)^{2}}{x-9} \geqslant 0 \Rightarrow \frac{x^{2}}{x-9}\geq36 $$ Therefore it attains its minimum 36 when $x=18.$

Lai
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