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