A simple inequality with ellipses

[RoNN|3]

Assume:

$$p>1, x>0, y>0$$

$$a \geq 1 \geq b > 0$$

$$\frac{a^2}{p^2}+(1-\frac{1}{p^2})b^2 \leq 1$$

$$\frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1$$


Prove:

$$\frac{x}{p}+y\sqrt{1-\frac{1}{p^2}} \leq 1$$


I've been trying for 3 days and it's driving me crazy. Any ideas?

 

[Stephen Tashi]

Have you been able to determine if there is a point on the ellipse
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$
where the tangent line is parallel to the line given by
$$\frac{x}{p}+y\sqrt{1-\frac{1}{p^2}} =1$$

That's probably not easy to do, but it looks like the most straightforward approach.

 

[RoNN|3]

Thank you very much. That approach works well. I am too tired/lazy to write the details here. If someone wants them, let me know.