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I know that this can be solved with calculus by differentiating w.r.t. x and equating it to zero. So, $f'(x)=(x^{p+q}-1)/x^{q+1}$ and on solving we find that $f(x)$is minimum at $x=1$.

But, I was wondering if there was a way of solving this problem by using the Holder's inequality as the conditions given in the question seem very similar to the conditions of the holder's inequality. Any other way of solving the problem without the use of calculus is most welcome. Thanks!

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    $\begingroup$ Unless I'm missing something, $f'(x) = x^{p-1} - x^{q-1}$ $\endgroup$ Sep 8 at 14:14
  • $\begingroup$ @TomCollinge Yea sorry i fixed it $\endgroup$ Sep 8 at 14:18
  • $\begingroup$ Not fixed yet.... $-1/x^{q+1} = x^{-q-1}$ $\endgroup$ Sep 8 at 14:19
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    $\begingroup$ It is an immediate application of Young's inequality for products $\endgroup$
    – Martin R
    Sep 8 at 14:41
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    $\begingroup$ @TomCollinge: Young's inequality with $a=x$ and $b=1/x$ ... $\endgroup$
    – Martin R
    Sep 8 at 14:44
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For $x > 0$ (otherwise the terms $x^p$ and $x^{-q}$ are not well-defined) we can apply Young's inequality for products:

$$ f(x) =\frac{x^p}{p}+\frac{(1/x)^{q}}{q} \ge x \frac 1x = 1 \, . $$ Equality holds if $x^p = (1/x)^q$, that is for $x=1$.

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