A simple inequality with ellipses

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The discussion centers on proving the inequality \(\frac{x}{p}+y\sqrt{1-\frac{1}{p^2}} \leq 1\) under the conditions \(p>1\), \(x>0\), \(y>0\), and \(a \geq 1 \geq b > 0\). Participants explore the relationship between the ellipse defined by \(\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\) and the line \(\frac{x}{p}+y\sqrt{1-\frac{1}{p^2}} =1\). A suggested approach involves finding a point on the ellipse where the tangent line is parallel to the given line, although the details of this method are not fully elaborated in the discussion.

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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?
 
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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.
 
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.
 

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