# Proving simple inequality

I have a simple function and want to make sure that it satisfies an inequality over a certain range of parameters.

My approach is:

In[125]:= a[x] := x^2

In[126]:= Reduce[a[x] > 0 && 1 <= x <= 2]

Out[126]= 1 <= x <= 2

However, I would expect an output of "True", since the inequality is satisfied over the entire range.

What do I need to invoke in order to obtain "True" as a result?

Thanks!

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• Reduce solves the system of inequalities and returns the range of x in which the system is true. this code returns True Refine[Reduce[a[x] > 0, x], 1 <= x <= 2] – k_v Apr 2 '15 at 8:57

Reduce is designed to reduce a set of inequalities to obtain the relevant information. In your case.

In:

Reduce[x^2 > 0 && 1 <= x <= 2]

Out:

1 <= x <= 2

tells you that your first inequality x^2>0 was an obsolete infomation, for the set of inequalities. In other words this tells you, that the first equation is allways satisfied if the second equation is satisfied.

So its a problem of interpretation here.

If you want to have an explicit test, that yields True or False, you could simply use this:

satisfiesInequality[ineq_, range_] :=
Reduce[ineq && range] === Reduce[range]

another approach would be to integrate the Boole-Function over the range where you want to test the first inequation.

satisfiesInequality2[
ineq_, {x_, xmin_, xmax_}] := (1 ===
Integrate[Boole[ineq], {x, xmin, xmax}]/(xmax - xmin))

add some SyntaxInformation that helps you using the function correctly:

SyntaxInformation[
satisfiesInequality2] = {"LocalVariables" -> {"Plot", {2,
Infinity}}, "ArgumentsPattern" -> {_, {_, _, _}}};

Show that $x^2 > 0$ for all $1\leq x\leq 2$.
$\forall_{x\in\left[1,2\right]}\;x^2>0$