Reduce[ForAll[{a, b}, {{a, b} > 1,
2 (a + b) == E^(2 a) + 2 Log[b] + 1}, #]] & /@ {1 < b < a,
a < b < 2 a, 2 a < b < E^a, E^a < b < E^(2 a)}
The code returned an error when solving this problem. Any suggestions for a better approach?
Maybe expand the list inequality (second part of ForAll construct)?
The syntax is correct, but this method cannot solve the problem.
We can find a numerical solution and then test the four conditions with it:
eq = 2 (a + b) == E^(2 a) + 2 Log[b] + 1;
conds = {1 < b < a, a < b < 2 a,
2 a < b < E^a, E^a < b < E^(2 a)};
conds /. FindInstance[eq && a > 1 && b > 1, {a, b}][[1]]
The output is
{False, False, False, True}
which means that only condition D has a chance.
To prove that condition D is true, we can call f[a,b] the difference between the two sides: this continuous function is monotonically increasing with respect to b, it is negative at b=Exp[a] and positive at b=Exp[2a]:
f[a_, b_] = 2 (a + b) - (E^(2 a) + 2 Log[b] + 1);
D[f[a, b], b] // Factor
f[a, Exp[a]] // PowerExpand // Simplify
f[a, Exp[2 a]] // PowerExpand // Simplify
Hence f vanishes in between.
Thank you very much! Using the FindInstance function to obtain a numerical solution resolved this issue. However, there are many similar types of problems, and a single unified method may not work for all of them—each case requires specific analysis.