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Question

if $n>m$, $\frac {a}{x^m} > 0$ and $x^{n-m} > 0$,prove $y= \frac {x^n+a}{x^m}$ is minimum when $x= \sqrt[n]{\frac {am}{n-m}}$ and value of minimum is equal to $y= \frac{n}{m}\sqrt[n]{(\frac {am}{n-m})^{n-m}}$

My idea

i know that if $P=x^n\times y^m\times...\times t^l$ is constant,then $S=x+y+...+t$ is minimum when $$\frac{x}{n}=\frac{y}{m}=...=\frac{t}{l}$$

and $y= \frac {x^n+a}{x^m}=x^{n-m}+ \frac {a} {x^m}$

user2838619
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2 Answers2

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Hint: $$\left(x^{n-m}\right)^{\frac1{n-m}}\cdot\left(\frac a{x^m}\right)^{\frac1m}=a^{\frac1m}$$ is certainly constant.

Bart Michels
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Another approach would involve derivatives. To find an extrema of a function we need to take its derivative and find its zeros: $$y'={nx^{n-1}x^m-mx^{m-1}(x^n+a)\over x^{2m}}\\ nx^{n-1}x^m-mx^{m-1}(x^n+a)=0\\ x^{m-1}(nx^n-mx^n-am)=0\\ x^n={ma\over n-m}\\ x_0=\sqrt[n]{{ma\over n-m}}\\ $$

We are still not done and need to prove that $x_0$ is a minimum. I leave it for you, but if you'd like me to show the solution, I will do it.

Dmitry Kazakov
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  • thanks for your solution.but even the book itself mentioned that there are ways to find minimum and maximum values with derivative but book don't want to use them because it's on inequality chapter. – user2838619 Jul 25 '14 at 09:19
  • Ok, next time it would be better if you mention such details in your question :) – Dmitry Kazakov Jul 25 '14 at 09:20
  • well i just tagged inequality.if i wanted derivatives, i would have included calculus tag. – user2838619 Jul 25 '14 at 09:21
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    Note that it can only be a minimum because 1) The function is $C^\infty$ on $]0,\infty[$. 2) $y'=0$ only once. 3) if $x\to0$ or $x\to\infty$ we have $y\to\infty$. @user2838619 we're happy to hear that you tag your questions carefully, but unfortunately not every user would do that, which makes we can't let our answers be influenced by the tagging. It is recommended to mention which methods you prefer. – Bart Michels Jul 25 '14 at 09:22
  • @barto nice, thank you! – Dmitry Kazakov Jul 25 '14 at 09:24