# Identifying the sign of an expression in an interval

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

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## migrated from math.stackexchange.comJul 20 '12 at 13:06

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Can you reproduce here your expression in Mathematica? – Andrew Jul 20 '12 at 12:23
I am sorry but I cannot understand what do you want me to do . Do you want to see what I am doing in Mathematica? My problem is that I just need to check whether the above expression is positive always. In case that it is not i can conclude that my results are ambiguous and say that the sign depends on the parameters. But first I need to know for sure that there is no way to find the sign. I can also assume that alpha is equal to 1/2 which I try the last hour using the following: – Kostas Jul 20 '12 at 12:54
:n = ((μhα)/(1 + h))^α ((μh*(1 - α))/(1 + h))^(1 -α) :q = ((μhα)/(1 + hα))^α ((μh*(1 - α))/((1 + hα)*(1 + h)))^(1-α) + (rμh^2*α*(2 + h)*(1 - α))/((1 + h*(1 - α))*(1 + h)*(1 + h*α)) :α = 1/2 :t = Assuming[h > 0 && r > 0 && μ > 0, Refine[q -n]] :Reduce[t >= 0, {h, r, μ}, Reals] – Kostas Jul 20 '12 at 12:55
Thank you for your replay, by the way – Kostas Jul 20 '12 at 12:59

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 $0<\alpha<1$.

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