# One inequality with two variables

I would just like to know if this:

((a - b)/(b*Log[b])) - Log[Log[a]] + Log[Log[b]]


exceeds 0 for any combination of a and b, being both variables positive integers.

I have tried to compute it on Mathematica but I only get hundreds of errors.

What would be the way to know if there is a solution to

((a - b)/(b*Log[b])) - Log[Log[a]] + Log[Log[b]]<0

• Welcome to Mathematica.SE! 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – user9660 Nov 17 '16 at 14:20
• Hmmm ... difficult, you could try FindInstance[((a - b)/(b*Log[b])) - Log[Log[a]] + Log[Log[b]] < 0 && a > 0 && b > b, {a, b}, Integers] but Mathematica does not give me any instance. Did you take a look at 3D plots of the function? – Mauricio Fernández Nov 17 '16 at 14:22

A mathematical proof would be best, but Mathematica can provide some evidence against the statement that

f[a_, b_] := ((a - b)/(b*Log[b])) - Log[Log[a]] + Log[Log[b]]


is negative for any integer a, b.

A straightforward way would be to do

NMinimize[{f[a, b], a > 1, b > 1}, {a, b}, Integers]


{0., {a -> 4, b -> 4}}

Indeed, f[a, b]==0 for a == b. Dropping the condition Integers,

NMinimize[{f[a, b], a > 1, b > 1}, {a, b}]


{-1.11022*10^-16, {a -> 3.71536, b -> 3.71536}}

gives a numerical zero.

A ContourPlot:

ContourPlot[f[a, b], {a, 1, 10}, {b, 1, 10}, PlotLegends -> Automatic] hints that indeed there are no truly negative values of f[a, b] for a, b >= 1.

A contour highlighting f[a, b]==0:

ContourPlot[f[a, b], {a, 1, 10}, {b, 1, 10}, Contours -> {0}] confirms that 0 is the smallest value attained by f[a, b] for positive integer a, b.

eq = ((a - b)/(b*Log[b])) - Log[Log[a]] + Log[Log[b]];


First try with FindInstance

FindInstance[eq < 0, {a, b}, Integers]


FindInstance::nsmet: The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist.

Second try:

data = Flatten[Table[{a, b, eq}, {a, 2, 100}, {b, 2, 100}], 1] // N;
MinimalBy[data, Last] probably never less than zero.

For this case you can try a graphical method. For example:

RegionPlot[((a - b)/(b*Log[b])) - Log[Log[a]] + Log[Log[b]] > 0, {a,
0, 10}, {b, 0, 10}]


It shows the region on the ab plane where the inequality is satisfied. Of course you can change the limits for a and b according to your needs. I've got a>1 and b>1 as a solution.