Prove that if $|a| \gt |b|$ then the ellipse
$\frac {x²}{a²}+\frac {y²}{b²}=1$
is contained within both the disk $x²+y²\le a²$ and the rectangle $|x|\le a,|y|\le b$
geometrically it is easy to see that this holds true, but I'm out of ideas of how to actually prove it algebrically
Let $(x,y)$ be a point on the ellipse $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ and assume $|a| > |b|$. We can conclude that $a^2 > b^2$ and then
$$ x^2 + y^2 \le x^2 + \frac{a^2}{b^2}y^2 = a^2 \tag1$$
Thus, $(x,y)$ is in the disk $x^2 + y^2 \le a^2$. Now for the rectangle,
$$ \frac{x^2}{a^2} = 1-\frac{y^2}{b^2} \le 1 \implies x^2 \le a^2 \implies |x| \le a $$
And similarly for $y$
$$ \frac{y^2}{b^2} = 1-\frac{x^2}{a^2} \le 1 \implies y^2 \le b^2 \implies |y| \le b $$
Note that for the rectangle we have not used the fact that $|a| > |b|$ and it is true in general. In addition, note that $(1)$ is not a strict inequality because $y$ may be $0$. Thus, we do not "gain" much by the strict restriction $|a|>|b| $ and we may as well require $|a|\ge |b|$ to get the same result.