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I want to show that $9*\left[\frac{xy}{x+y}-q(1-q)\right]-12*[xy-q(1-q)]+(1-q-x)^{3}+(x+y)^{3}+(q-y)^{3}-1\geq0$ where

$0<q<1$

$0<x<1-q$

$0<y<q$

$(x+y)\left[1+max\{\frac{1-q}{y},\frac{q}{x}\}\right]\leq3$

I play with it numerically. It is right. But don't know how to prove it analytically. Anybody can help? Thanks a lot.

Peter
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    Our general advice is that your post title should say something about the subject of the question. Mentioning that it's an inequality is good, but that it came up in your research is irrelevant for someone deciding whether to visit it. Can you say anything more informative? – LSpice May 03 '19 at 01:39

3 Answers3

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Iosif's answer is very interesting and if anyone knows how to do the same thing in Maple I'd like to know.

However I disagree with Iosif about the difficulty. Mathematica will use a systematic procedure that is guaranteed to work in a wide variety of cases, and that may take many more steps than an ad hoc method devised by a human.

So here is a quick proof. Remove the upper bounds on $q$, $x$, $y$ by substituting $x=(1-q)X/(1+X)$, $y=qY/(1+Y)$, $q=Q/(1+Q)$. After clearing demoninators that are obviously positive, we have to prove $\Phi\ge 0$ subject to $Q,X,Y\ge 0$ and conditions $C_1$ and $C_2$, where $$\begin{align}\Phi &= -Y^2(1+X)^3 Q^4 + XY(1+X)^2(Y+2)Q^3\\ &{\quad}+ 2XY(1+Y)(1+X)(X+Y+3)Q^2 + XY(1+Y)^2(X+2)Q - X^2(1+Y)^3\end{align}$$ and $C_1$, $C_2$ come from the OP's last condition. They can be arranged like this: \begin{align*} C_1:\qquad & YQ \ge \frac{(1+Y)X}{(X+2)}-\frac{Y(2Y+3)(1+X)}{(1+Y)(X+2)}Q^2. \\ C_2:\qquad & YQ^2 \le \frac{X(1+Y)(2X+3)}{(1+X)^2}+\frac{(Y+2)X}{(1+X)}Q. \end{align*}

Now apply $C_1$ to the linear term of $\Phi$ and apply $C_2$ to the quartic term in the manner $YQ^4 \le \operatorname{rhs}(C_2)\,Q^2$. The result is exactly 0.

To show that $\Phi\gt 0$ strictly when $x,y,q\gt 0$, note that zero can only happen if $C_1,C_2$ hold with equality at the same time. But $Q\operatorname{rhs}(C_1)-\operatorname{rhs}(C_2)$ is manifestly negative.

LSpice
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Brendan McKay
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  • All right. I have not checked your proof, but it looks like it required quite a bit of ingenuity. Does ingenuity count towards difficulty? :-) Can you explain how you came up with your applications of $C_1$ and $C_2$? – Iosif Pinelis May 03 '19 at 13:21
  • @IosifPinelis Looking at $\Phi$, only the constant term and the quartic term can be negative. From the signs of the coefficients of $C_1$, it seemed that the linear term could be sacrificed to eliminate the constant term. To eliminate the quartic term using $C_2$, I first tried $Y^2Q^4\ge \operatorname{rhs}(C_2)^2$ but that didn't work, so I tried what is written. I was surprised to get exactly 0, but that could reflect the source of the inequality. – Brendan McKay May 03 '19 at 14:49
  • ... I meant $Y^2Q^4 \le \operatorname{rhs}(C_2)^2$ of course. – Brendan McKay May 03 '19 at 14:59
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    The trick is really fantastic! Thank you so much. I manage to verify everything except that I have a hard time deriving $\Phi$. Do you use a software to obtain such a concise expression? – Peter May 03 '19 at 18:08
  • Finally, I derive $\Phi$ using Mathematica. Confirm your answer is correct. Thank you – Peter May 03 '19 at 18:31
  • When I was young I would have done it by hand and got the right answer, but now I make too many mistakes and rely on my computer for routine things. – Brendan McKay May 04 '19 at 05:36
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Welcome to MO! However, your conjecture is false e.g. for $q = 1/2$, $x= 3/8$, $y = 1/4$, $t= 0$.

Added: The OP later stated that the additional condition $t=3$ was initially omitted in the OP's post. Anyhow, the problem is one of real algebraic geometry and, as such, admits a completely algorithmic solution. In Mathematica, such algorithms are represented by Reduce[] and related commands. Using Reduce[] indeed, we get

enter image description here

This proves the conjecture. We see that it took Mathematica about 2 sec to obtain this result; so, a manual proof might be quite long and laborious, and most likely less reliable than Mathematica's.

Iosif Pinelis
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  • Thank you so much! Your answer is extremely helpful. Can I obtain the proof from Mathematica? I need to attach a proof to my paper. I am not sure whether people think it is valid if I claim Mathematica can prove it. – Peter May 03 '19 at 04:00
  • @Peter : What happens in Mathematica's innards is not available to general public. Even if it were available, that proof may take tens or even hundreds of pages -- think how much processing Mathematica can do in 2 sec. Computer-assisted proofs have already gained a great deal of (well-deserved) acceptance, and I actually have several published papers containing such proofs. However, since Mathematica is proprietary, you may try instead to use such free computer algebra systems as Reduce (https://reduce-algebra.sourceforge.io/ ). In my experience, they are not as good as Mathematica, though. – Iosif Pinelis May 03 '19 at 04:13
  • Previous comment, continued: You may want to attach a proof such as the above one to your paper or maybe a link to it. I have done both such things. – Iosif Pinelis May 03 '19 at 04:17
  • Sure. Given the problem takes Mathematica two seconds, attaching the mathematica code and results seems to be only possible way. Thank you again. I have played with the problem for two whole days. – Peter May 03 '19 at 05:12
  • @IosifPinelis Please see my answer, as I think it shows it is a mistake to infer that a problem is difficult on account of Mathematica taking a long time. – Brendan McKay May 03 '19 at 07:19
  • @BrendanMcKay : Of course, I agree that, because Mathematica's algorithms are generic, sometimes a human can outperform Mathematica. On a few occasions, I have done it myself; see e.g. the answer at https://mathoverflow.net/questions/239243/an-inequality-concerning-lagranges-identity, where Mathematica took 22 min (after some help by a human), but a rather short proof was eventually found. – Iosif Pinelis May 03 '19 at 13:01
  • Previous comment continued: However, I did not say (or think) that, if Mathematica takes a significant amount of time on solving a problem, then it is necessarily difficult. What I said was merely that, to transcribe Mathematica's 2-sec proof into (say) English, it may take tens or even hundreds of pages. – Iosif Pinelis May 03 '19 at 13:01
  • I have a question about the algorithmic approach to verifying this inequality. Could something similar be done if the terms also involved logarithms? I have a problem that would be finished if I could establish a tricky but elementary inequality involving rational functions and logarithms. Is there a Mathematica approach to this or can that only be done in the context of algebraic geometry? – Gabe K May 03 '19 at 18:54
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    @GabeK It is possible to use interval arithmetic to prove inequalities which are not just polynomials, but more generally non-linear expressions. For instance, one can prove inequalities for the expression $2^{-2-3p} (2^{3 p} + 5\cdot 2^{2 + p} + 12 (3- \sqrt{5})^p + 12 (3 + \sqrt{5})^p)$ on an interval using the interval arithmetic library in PVS. It is not clear if the 'Reduce[]' command in Mathematica uses this approach to handle such functions, or other algebro-geometric methods. – Josiah Park May 03 '19 at 21:13
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    @GabeK : It is possible to reduce many problems involving logarithms to algebraic problems, since the derivative of $\ln$ is algebraic. Mathematica's Reduce[] can indeed solve some problems involving logarithms. – Iosif Pinelis May 03 '19 at 22:13
  • Thanks. I'll have to see if that works. I appreciate it. – Gabe K May 04 '19 at 03:45
  • Dear @IosifPinelis: Would you be so kind as to take a look at this inequality https://math.stackexchange.com/q/3335271/64809? Thank you! – Hans Aug 26 '19 at 21:20
  • @Hans : My activities on MathSE are very marginal. Can you re-post this problem on MO? – Iosif Pinelis Aug 27 '19 at 01:51
  • @Hans : I have, however, posted an answer on MathSE. – Iosif Pinelis Aug 27 '19 at 02:16
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Remark: Actually, the inequality is quite easy. The trick is to take $x, y$ as parameters.

Denote the expression by $f(q)$.

It suffices to prove that $f(q) \ge 0$ provided that \begin{align*} &x, y > 0,\\ &x + y < 1,\\ &3y > (x + y)^2,\\ &3x > (x + y)^2, \\ &y < q < 1 - x, \\ &1 + y - \frac{3y}{x + y} \le q \le \frac{3x}{x + y} - x. \end{align*} (Note: Condition $(x+y)\left[1+max\{\frac{1-q}{y},\frac{q}{x}\}\right]\leq3$ is equivalent to $(x + y)[1 + (1-q)/y] \le 3$ and $(x + y)(1 + q/x) \le 3$ which are equivalent to $1 + y - \frac{3y}{x + y} \le q \le \frac{3x}{x + y} - x$. Also, from $y < \frac{3x}{x + y} - x$, we have $3x > (x + y)^2$, and from $1 + y - \frac{3y}{x + y} < 1 - x $, we have $3y > (x + y)^2$.)

Note that $f(q)$ is quadratic and concave (the coefficient of $q^2$ is $-3x - 3y$).

Also, we have \begin{align*} f(y) &= \frac{3x(1 - x - y)(2y - x)}{x + y},\\ f(1 - x) &= \frac{3y(1 - x - y)(2x - y)}{x + y},\\ f\left(1 + y - \frac{3y}{x + y}\right) &= \frac{3y(2 - x - y)(x - 2y)}{x + y}, \\ f\left(\frac{3x}{x + y} - x\right) &= \frac{3x(2 - x - y)(y - 2x)}{x + y}. \end{align*}

We split into three cases:

(1) If $x > 2y$, we have $1 + y - \frac{3y}{x + y} > y$ and $1 - x \le \frac{3x}{x + y} - x$, and $f(1 + y - \frac{3y}{x + y}) \ge 0$ and $f(1 - x) \ge 0$. Thus, $f(q) \ge 0$.

(2) If $y/2 \le x \le 2y$, we have $1 + y - \frac{3y}{x + y} \le y$ and $1 - x \le \frac{3x}{x + y} - x$, and $f(y) \ge 0$ and $f(1 - x) \ge 0$. Thus, we have $f(q) \ge 0$.

(3) If $x < y/2$, we have $1 + y - \frac{3y}{x + y} \le y$ and $1 - x > \frac{3x}{x + y} - x$, and $f(y) \ge 0$ and $f(\frac{3x}{x + y} - x) \ge 0$. Thus, $f(q) \ge 0$.

We are done.

River Li
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