Show Inequality Problem: 0 < x < 1, 0 < y < x/√3 + 1/√3

[blackbelt5400]

Homework Statement

Show that $$\frac{2x+1}{[(x+1)^2+y^2]^{3/2}}+\frac{x-1}{[(x-1)^2+y^2]^{3/2}} < 0$$ for $$0 < x < 1$$ and $$0 < y < \frac{x}{\sqrt{3}}+\frac{1}{\sqrt{3}}$$.

Homework Equations

The Attempt at a Solution

I've confirmed by graphing in Maple.

It's easy to see that $$(x+1)^2+y^2 > (x-1)^2+y^2$$, and therefore $$\frac{2x+1}{[(x+1)^2+y^2]^{3/2}}<\frac{2x+1}{[(x-1)^2+y^2]^{3/2}}$$. Unfortunately, this is a dead end because when you add the two fractions, the numerator becomes 3x, which is strictly positive.

I've tried differentiating both terms with respect to x to show that the negative term is decreasing faster than the positive term; since at x=0 the sum is 0, this would imply that the sum is always negative for x>0, but I've had no luck.

 

[Defennder]

I can think of a tedious way to do this: find the critical points, then perform the second derivative test in 2 variables (using the Hessian) to see if you can find the local maximum. If there is a local maximum, it should be lesser than 0. But the partial derivatives for both x,y is kind of hellish to compute, so I suppose there has to be a better way.

 

[Dick]

That can't be the actual problem you were given, is it? With no motivation for solving it? It must mean something, yes?

 

[blackbelt5400]

I can think of a tedious way to do this: find the critical points, then perform the second derivative test in 2 variables (using the Hessian) to see if you can find the local maximum. If there is a local maximum, it should be lesser than 0. But the partial derivatives for both x,y is kind of hellish to compute, so I suppose there has to be a better way.

Unfortunately, I already know that at no point of the given region will I have critical points, so this method would not be useful. I unfortunately have been down this path before with a similar problem.

 

[blackbelt5400]

That can't be the actual problem you were given, is it? With no motivation for solving it? It must mean something, yes?

Yes, there is an actual problem here. This is a small portion of one of the proofs for a theorem in my thesis on electrostatics and potential theory.

Consider a configuration of three equal, positive point charges arranged in an equilateral triangle, with the charges located at $$(-1,0), (1,0),$$ and $$(0, \sqrt{3})$$. Define the electrostatic potential using the standard Newtonian kernel; that is:

$$P(x,y)=\sum_{k=1}^3{\frac{q_k}{\sqrt{(x-x_k)^2+(y-y_k)^2}}}$$

The electrostatic force is given by the gradient of the potential: $$F= -\nabla P$$

In particular, the x-component of the force is
$$F_x = -\frac{\partial P}{\partial x} = \frac{x}{[x^2+(y-\sqrt{3})^2]^{3/2}}+\frac{x+1}{[(x+1)^2+y^2]^{3/2}}+\frac{x-1}{[(x-1)^2+y^2]^{3/2}}$$

I have already proven that the electrostatic force vanishes at four points: all on the three altitudes of the equilateral triangle. The goal is to prove that there are not more than 4 equilibrium points. Using the three altitudes, you can partition the equilateral triangle into six right triangles, one of which is determined by 0<x<1 and $$0<y<-\frac{x}{\sqrt{3}}+\frac{1}{\sqrt{3}}$$. The goal is to show that the x-component of the force vectors in this triangle always point to the left... i.e. Fx < 0. If this can be established, then by symmetry I can eliminate any equilibrium points from the other 5 right triangles, leaving me with the four which I have established.

Notice that the first two terms of Fx are positive. I was able to distill them together into the form $$\frac{2x+1}{[(x+1)^2+y^2]^{3/2}}$$. What remains is to show that the resulting sum is negative.

 

[blackbelt5400]

I may have answered my own question, but I don't trust my results at 2 am. Does this make sense?

Claim: $$\frac{2x+1}{(x+1)^2+y^2}<\frac{1-x}{[(x-1)^2+y^2]^{3/2}}.$$

Proof of Claim: Notice that along the line x=0, equality holds. However, we are only considering 0<x<1; also, notice that this bound on x gives that $$0<y<\frac{1}{\sqrt{3}}.$$ It is trivial to see that $$(x+1)^2+y^2>(x-1)^2+y^2.$$ Applying these inequalities gives $$\begin{eqnarray*} \frac{\partial}{\partial x}(\frac{2x+1}{[(x+1)^2+y^2]^{3/2}}) &=&\frac{-4x^2-5x-1+2y^2}{[(x+1)^2+y^2]^{5/2}}\\ &<&\frac{-4x^2-5x-\frac{1}{3}}{[(x+1)^2+y^2]^{5/2}}\\ &<&\frac{2x^2-4x+2-\frac{1}{3}}{[(x+1)^2+y^2]^{5/2}}\\ &<&\frac{2x^2-4x+2-y^2}{[(x+1)^2+y^2]^{5/2}}\\ &<&\frac{2(x-1)^2-y^2}{[(x-1)^2+y^2]^{5/2}}\\ &=&\frac{\partial}{\partial x}(\frac{1-x}{[(x-1)^2+y^2]^{3/2}}). \end{eqnarray*}$$
Since $$\frac{\partial}{\partial x}(\frac{2x+1}{[(x+1)^2+y^2]^{3/2}})<\frac{\partial}{\partial x}(\frac{1-x}{[(x-1)^2+y^2]^{3/2}})$$, and $$\frac{2x+1}{[(x+1)^2+y^2]^{3/2}}=\frac{1-x}{[(x-1)^2+y^2]^{3/2}}$$ when x=0, then at no point inside the triangle can equiality hold. Therefore $$\frac{2x+1}{[(x+1)^2+y^2]^{3/2}}<\frac{1-x}{[(x-1)^2+y^2]^{3/2}}$$, and the Claim is proven.

Substituting the result of this claim back into the original problem shows that the sum is negative, correct?


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