Conjecture on the Minimum value of $a\log(bc) + bc \log(a)$.

Conjecture: Given three successive odd primes, the value of the above expression is minimum if the largest of the three odd primes is set as a, and the remaining two are set as b and c.

Here is a solution sent to me by a kind and very famous mathematician:

The rule you observed “ pick the largest of a, b, c holds as soon as a, b, c are relatively close and not too small, which holds for successive odd primes. For the three, (x, y)= (a, bc), (b, ca) or (c, ab) you have the same value of xy. For X=log x, Y=log y , we have the same value for X+Y=S= log (abc) For X + Y= fixed S, not too small you consider: (e^X).Y+ X.(e^Y) (where X+Y=S) As a function of X in [0.S], the graph of this function looks like:(symmetrical in X<->S-X) and you are near X~S/3, there it is decreasing. Hence your rule.

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  • $\begingroup$ Do you mean $a\log(bc)+bc\log(a)$? $\endgroup$ – Thomas Andrews Jul 15 '17 at 0:32
  • $\begingroup$ yes Tom, thanks for the correction. $\endgroup$ – prashanth rao Jul 15 '17 at 0:33
  • $\begingroup$ just making sure, log is the base 10 log ? $\endgroup$ – user451844 Jul 15 '17 at 1:04
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    $\begingroup$ Base doesn't matter (as long as the base is $>1$) since the above can be written as $\log(b^ac^aa^{bc})$. So you are really trying to minimize $b^ac^aa^{bc}$. @RoddyMacPhee $\endgroup$ – Thomas Andrews Jul 15 '17 at 1:13
  • $\begingroup$ So... you are asking if $b, c < a$ to prove $a \log bc + bc \log a$ is less than either $b\log ac + ac \log b$ or $c\log ab + ab \log c$? $\endgroup$ – fleablood Jul 15 '17 at 1:22

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