Inequality $x+y+xy\geq c$

Question

I have two non-negative numbers satisfying: $$ x+y+xy\geq c>0 $$ Can I find a $\delta(c)>0$ such that $$ x^2+y^2\geq\delta(c) $$ It seems obvious but I can not determine $\delta(c)$ explicitly?

Answer 1

Suppose $x^2+y^2 \le k$. In particular, $x,y \le \sqrt{k}$. So $$ x+y+xy \le k+\sqrt{k} $$ which is impossible if $k$ is too small.

Answer 2

$${x^2+y^2 \over 2} +{x^2+1\over 2 }+{y^2+1\over 2} \geq xy+x+y\geq c$$

so we have $$x^2+y^2\geq 2c-2$$

Answer 3

You have $(x+1)^2+(y+1)^2 = x^2+y^2+2(x+y)+2\geq 2(x+1)(y+1)\geq 2c+2$ and also $x+y\leq \sqrt{2x^2+2y^2}$. So, if you let $x^2+y^2 = a^2$, then it follows that $$a^2+2\sqrt{2}a\geq 2c$$ and then you can solve this to get a lower bound for $a.$

Answer 4

Partial solution: If we assume the curve $y = \frac{c-x}{1+x}$ and the circle $x^2 + y^2 = r^2$ have a common tangent at some point. We need to find the smallest such radius $r$.

Need to minimize: $x^2 + y^2 = x^2 + \frac{(c-x)^2}{(1+x)^2}$

Differentiate to solve for $x$ in terms of $c$. Let this be $x_1$

Then $\delta(c) = \left({x_1^2 + \frac{(c-x_1)^2}{(1+x_1)^2}}\right)^{\frac{1}{2}}$

Not sure if there is an elegant approach.