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Let $x\in(0,1)$,and $-1<p-q<0,p>1,p,q\in\mathbb R$. Prove or disprove $$q(q-1)x^q-p(p-1)x^p\ge 0.$$

sranthrop
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2 Answers2

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The inequality is false. Indeed, if it were true then $$ q(q-1)x^q\geq p(p-1)x^p $$ would imply that $q\log x + \log q + \log (q-1)$ is an increasing function of $q$, for any fixed value of $x\in (0,1)$ and for $q>1$.

In this domain, the function is differentiable and $$ \frac{d}{dq}\left[q\log x + \log q + \log (q-1)\right]=\log x+\frac{1}{q}+\frac{1}{q-1}. $$ As $q\to\infty$ this expression tends to $\log x<0$, hence for all sufficiently large $q$ the expression is negative. Thus $q\log x+\log q +\log(q-1)$ is not an increasing function, so the inequality is not true.

pre-kidney
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  • I don't follow. $$ q(q-1)x^q\geq p(p-1)x^p $$ may also imply that LHS is a decreasing function, so is RHS, but at all times, LHS >= RHS. How is this handled? – Deepak Gupta Sep 22 '15 at 06:33
  • Edit: Kindly ignore my previous comment. I understand the solution now. Elegant! Thanks! This can be used to generate two examples - one where the inequality holds and the other where it does not. – Deepak Gupta Sep 22 '15 at 06:59
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The inequality can't be true in general. It is equivalent to \begin{align*} x^{q-p}\geq\frac{p(p-1)}{q(q-1)}. \end{align*} Note that $p-q<0$ and $p>1$ implies that $q-p>0$ and $\frac{p(p-1)}{q(q-1)}>0$. But if $x\to0^+$, then $x^{q-p}\to0$. To be more precise, the inequality is true, if and only if \begin{align*} x\geq\left(\frac{p(p-1)}{q(q-1)}\right)^\frac{1}{q-p}. \end{align*}

sranthrop
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  • This is not adequate in the scenario $q \to p^-$, in which case LHS takes the form $0^0$ as $x \to 0^+$. In fact, $\lim_{q \to p^-} {x^{q-p}} = 1 = \lim_{q \to p^-}{\frac {q(q-1)}{p(p-1)}} = 1$ – Deepak Gupta Sep 22 '15 at 06:40
  • No. Unless I misunderstood Thailandasw, $p,q,x$ are fixed real numbers, and I showed that you can always construct a counterexample by picking a fixed! $x$ sufficiently close to 0. Moreover, the case $0^0$ only occurs if the limits $x\to0^+$ and $q\to p^-$ are taken simultanously, which is not the case here. – sranthrop Sep 22 '15 at 06:43
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    The way the question is worded, $p,q,x$ are not fixed. The way to go ahead here would be to generate two examples with two sets of values $p_1, q_1, x_1$ and $p_2, q_2, x_2$ such that the inequality holds in one case but does not in the other. That would be conclusive in proving the inequality to be not strictly applicable. I guess you did find a case where the inequality does not hold which is $x \to 0^+, q \to p+1$ and I guess the counterexample could be proven when $x \to 1^-, q \to p^+$. But the latter does not seem to be clear to me – Deepak Gupta Sep 22 '15 at 06:51
  • The way the question is worded, $p,q,x$ are clearly fixed, but arbitrary real numbers (satisfying the given constraints). I gave a proof that no matter how $p,q$ are picked, as long as they satisfy the given constraints, one can always find $x$ close to 0 such that the inequality does not hold. In particular, if you give me $p,q$, then I give you an x such that the inequality does not hold. – sranthrop Sep 22 '15 at 06:59
  • In conclusion, for each and every value of x, one can find p and q such that the given inequality holds/ does not hold. Also, for any particular p and q, there exists at least one x for which the inequality does not hold true. While both statements serve the purpose of answering this question, I guess these two statements are equivalent? – Deepak Gupta Sep 22 '15 at 07:12
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    Yes. I added one more detail to my answer hopefully making everything clearer to you now :) – sranthrop Sep 22 '15 at 07:14