# Math Help - Inequalities. Is there a solution?

1. ## Inequalities. Is there a solution?

Hi, I was working with some problem when I came up with the following inequalities. I want to determine whether a solution exists. How does one tackle this monstrous beast? Where can I start? Thanks!

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Edit: I didn't expect it when I post the thread, but maybe some calculus is necessary! If so, Admin please feel free to move the thread to calculus subforum.

2. ## Re: Inequalities. Is there a solution?

I generated tens of millions of $({x}_{1},{x}_{2},{x}_{3},{x}_{4})$~U(0,1), where U(0,1) is uniform distribution on (0,1), and then sorted the 4-tuple in ascending order to check whether any of them satisfies the inequalities. I run the program several times, and none does! So one might as well work in that direction... But how???

3. ## Re: Inequalities. Is there a solution?

why have you confined your simulations to the range (0,1) ?

4. ## Re: Inequalities. Is there a solution?

Originally Posted by SpringFan25
why have you confined your simulations to the range (0,1) ?
Notice that the inequalities are homogeneous! Hence (0,1) will suffice!

5. ## Re: Inequalities. Is there a solution?

then you can assign any number (eg x4=1) to one of the unknowns and simplify your problem somewhat, not sure how much that will help though.

6. ## Re: Inequalities. Is there a solution?

Thanks SF! Let me set x3=1 then. Please check if there's anything wrong with the following proof:

Proof: Let f(x1,x2,1,x4)=LHS of the first inequality, then
df(x1,x2,1,x4)/dx2= $3{x}_{4}+{x}_{4}^{2}+{x}_{1}-{x}_{1}{x}_{4}>0$

Hence, if the second inequality is satisfied, we must have f(x1,x2,1,x4)>f(x1,(x1+x4)/2,1,x4).

Now df(x1,(x1+x4)/2,1,x4)/dx1= $(1-{x}_{4}){x}_{1}+{x}_{4}-3{x}_{4}^2<0$. Hence f(x1,x2,1,x4)>f(x1,(x1+x4)/2,1,x4)>f(2-x4,(2-x4+x4)/2,1,x4)=f(2-x4,1,1,x4)=2(x4+1)(x4-1)^2>0, where the second inequality follows by substituting x1 with 2-x4, since 2x2>x1+x4 $\Rightarrow$ x1<2x2-x4<2x3-x4=2-x4. Hence the first inequality must be violated! Q.E.D.

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Edit: Red letters