About an inequality

Question

I have an inequality of the form

$$\frac{u^2+v^2}{2u}\leq 1$$

How one can determine the range for $u$ and $v$ to verify this condition.

Answer 1

If $u<0$, the equality is true for every value of $v$.

Now suppose $u > 0$. You can rewrite this inequality as

$$u^2 - 2u + v^2 \leq 0$$

Now, this is a polynomial of degree 2 in $u$, hence it's negative for

$$\frac{ 2 - \sqrt{ 4 - 4v^2 } }{2} \leq u \leq \frac{ 2 + \sqrt{ 4 - 4v^2 } }{2}$$

Note that you need $|v| \leq 1$ to have a solution