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Suppose that $0 \le {X_1} < {X_2} < {X_3}$ . How is it possible to prove the following function is increasing based on ${X_1}$ in the range of $0 \le {X_1} < {X_2}$ ?

$f({X_1},{X_2},{X_3}) = \frac{{(c + p{X_1}){{({X_2} - {X_3})}^2} + (c + p{X_2}){{({X_1} - {X_3})}^2} + (c + p{X_3}){{({X_1} - {X_2})}^2}}}{{{{[{X_1}{X_2}{\mkern 1mu} \log \frac{{{X_1}}}{{{X_2}}} + {X_1}{X_3}\log \frac{{{X_3}}}{{{X_1}}} + {X_2}{X_3}\log \frac{{{X_2}}}{{{X_3}}}]}^2}}}$

$c$ and $p$ are positive constants. I have plotted this function (as a function of ${X_1}$) and I observed that this is an increasing function. Then I tried to prove it analytically but taking derivative makes it too complicated, because I would come up with lots of conditions.

I also considered the numerator and denominator separately to investigate the behavior of each function, but it did not work. Because there was a contradiction in the common part of the conditions (ranges in which the derivative of numerator is positive and derivative of denominator is negative).

The most important point is that I should not put any constraint on ${X_1}$, but any condition is allowed to put on the ${X_2}$ and ${X_3}$.

Any kind of help would be so appreciated!

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  • $\begingroup$ If $c$ and $p$ are fixed, it would help to give their values. Or, if this is supposed to proved for all positive $c$ and $p$, then we can reduce to proving this in the case $c=0, \ p=1$ and the case $c=1,\ p=0$. $\endgroup$
    – user44143
    Commented Mar 21, 2019 at 17:09
  • $\begingroup$ Thank you @MattF. , I did it but it does not help with the solution. $\endgroup$
    – Hamed
    Commented Mar 27, 2019 at 18:49
  • $\begingroup$ What did you do? What are $c$ and $p$? $\endgroup$
    – user44143
    Commented Mar 27, 2019 at 19:42

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