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I am trying to use Mathematica to find the following maximum value:

$$\max_{0<\lambda< 1}\left| \frac{\lambda^{1-2p}}{(\lambda+\alpha)^2}\right|$$

where $0<p<1/2$ and $\alpha>0$

I used

Assuming[0 <= p <= 1/2 && \[Alpha] > 0, MaxValue[{\[Lambda]^(1 - 2*p)/(\[Lambda] + \[Alpha])^2, 
\[Lambda] > 0 && \[Lambda] != \[Alpha]}, \[Lambda]]]

I thought this would be quite easy, but I get the error code

FindMaxValue::eqineq: Constraints in {\[Lambda]>0,\[Lambda]!=\[Alpha]} are not all equality or inequality constraints. With the exception of integer domain constraints for linear programming, domain constraints or constraints with Unequal (!=) are not supported. >>

Note that I'd like to return an algebraic expression...so $\lambda=..$ something in terms of $\alpha$

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You could just do it like this (I have changed variable g=1-2p, a=α):

f[t_, a_, g_] := t^g/(t + a)^2

Find critical point(one in this case):

sol = Solve[D[f[t, a, g], t] == 0, t][[1]];

Second derivative test:

test = Refine[
  FullSimplify[D[f[t, a, g], {t, 2}] /. sol] < 0, {0 < g < 1, 
   a > 0, {a, g} \[Element] Reals}]

yields: True. The value of maximum:

val = FullSimplify[f[t, a, g] /. sol]

yields:

$\frac{(g-2)^2 \left(-\frac{a g}{g-2}\right)^g}{4 a^2}$

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  • $\begingroup$ Does sol belong to $(0,1)$? $\endgroup$ – user64494 Oct 24 '17 at 15:56

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