Identifying the sign of an expression in an interval

Question

My expression is the following: $$ \frac{(1-\alpha ) \alpha h^2 (h+2) \mu r}{(h+1) ((1-\alpha ) h+1) (\alpha h+1)}-\left(\frac{\alpha h \mu }{h+1}\right)^{\alpha } \left(\frac{(1-\alpha ) h \mu }{h+1}\right)^{1-\alpha }+\left(\frac{(1-\alpha ) h \mu }{(h+1) (\alpha h+1)}\right)^{1-\alpha } \left(\frac{\alpha h \mu }{\alpha h+1}\right)^{\alpha } $$ or given as Mathematica input:

(((μ*h*α)/(1 + h*α))^α ((μ*h*(1 - α))/((1 + h*α)*(1 + h)))^(1 -α)) + 
((r*μ*h^2*α*(2 + h)*(1-α))/((1 + h*(1 - α))*(1 + h)*(1 +h*α))) - 
(((μ*h*α)/(1 + h))^α ((μ*h*(1 - α))/(1 + h))^(1 -α))

and I would like to know the sign of this when : $1/2\leq α\leq 1$ and $h$,$r$ and $\mu$ are positive

Answer 1

Define

expr[mu_, h_, r_, 
   a_] := ((mu h a)/(1 + h a))^
    a ((mu h (1 - a))/((1 + h a) (1 + h)))^(1 - a) + (
   r mu h^2 a (2 + h) (1 - a))/((1 + h (1 - a)) (1 + h) (1 + 
      h a)) - ((mu h a)/(1 + h))^a ((mu h (1 - a))/(1 + h))^(1 - a);

Then the limits of $\alpha \downarrow 0$ and $\alpha \uparrow 1$ are easy for Mathematica:

In[15]:= Limit[expr[mu, h, r, a], a -> 0, Direction -> -1]

Out[15]= 0

In[16]:= Limit[expr[mu, h, r, a], a -> 1, Direction -> 1]

Out[16]= 0

Inspection of $\alpha=\frac{1}{2}$ shows that the expression can have either sign:

In[25]:= Reduce[expr[mu, h, r, 1/2] < 0 && h > 0 && mu > 0 && r > 0]

Out[25]= 0 < r < 1/2 && h > (8 r)/((1-2r)^2) && mu > 0

Playing with other values of $\alpha$ indicates that this remain true for all $0<\alpha<1$.