How to prove the solution of an inequality is lower bounded? Let $x_{n,k}\in (0,1]$ be any solution of the following inequality
\begin{align}
n(1-x)^2+2kx^2(1-x)^2 -x^2-k(1-x)^4 \leq 0,
\end{align}
where $n, k$ are parameters such that $n\geq k\geq 1$.
I want to prove that there exists a strictly positive constant $c$ (i.e. $c>0$) such that
\begin{align}
x_{n,k} \geq c, \text{ for all } n\geq k \geq 1.
\end{align}
If you want to have a look at their graphs, they are available here thanks to Wolfram Alpha.
I tried a lot of estimations, but failed.
I guess that we should use something special here.
Could someone give me a hint? Thanks!

Let $f(x) = n(1-x)^2+2kx^2(1-x)^2 -x^2-k(1-x)^4$.
We have, for all $x$ in $(0, 1]$,
\begin{align}
f(x) &\ge k(1-x)^2 + 0 - x^2 - k(1-x)^4 \\
&= k[(1-x)^2 - (1-x)^4] - x^2\\
&\ge (1-x)^2 - (1-x)^4 - x^2 \\
&\ge (1-x)^2 - (1-x)^3 - x^2 \\
&= x(x^2 - 3x + 1).
\end{align}
Thus, if $x\in (0, 1]$ satisfying $f(x) \le 0$, then
$x^2 - 3x + 1 \le 0$ which results in $\frac{3 - \sqrt{5}}{2} \le x \le 1$.
We are done.
