Let x,y > 0.
Let B>b>2.
Expb is exp base b and expB is exp base B.
Analogue for ln.
Conjecture :
A(x,y) = expB^[1/2](lnB^[1/2](x) + lnB^[1/2](y))
B(x,y) = expb^[1/2](lnb^[1/2](x) + lnb^[1/2](y))
C(x,y) = (2+x^2+y^2)^(B-b)
A(x,y)/( B(x,y) ln(2+C(x,y)) ) < 2
Regards
Tommy1729
Let B>b>2.
Expb is exp base b and expB is exp base B.
Analogue for ln.
Conjecture :
A(x,y) = expB^[1/2](lnB^[1/2](x) + lnB^[1/2](y))
B(x,y) = expb^[1/2](lnb^[1/2](x) + lnb^[1/2](y))
C(x,y) = (2+x^2+y^2)^(B-b)
A(x,y)/( B(x,y) ln(2+C(x,y)) ) < 2
Regards
Tommy1729