Maximize parameter $a$ such that inequality is still satisfied Consider the following inequality
$$ax^5 -bx^2 +2x -1 \leq 0$$
with the following constraints on the parameters $a,b$
$$a>0\\
b\in \left(0,\tfrac{1}{2}\right)\\$$
and the constraint on the variable $x$
$$x\in [1,\infty)$$
I want to do the following: For any given $b\in\left(0,\tfrac{1}{2}\right)$, find the largest $a>0$ such that there exists an $x\in[1,\infty)$ such that the inequality is satisfied. I now realize that this $x$ will then saturate the equality, so essentially: Given $b$, I am looking for the maximal $a$ such that the polynomial $ax^5 -bx^2 +2x -1$ still has roots $\geq$ 1.
For notational clarity: I am looking for the functional behavior of $a$ as a function of $b$ in the sense
$$a(b)= \{\max_{a>0}a \;|\;ax^5 -bx^2 +2x -1\leq0 , x\geq1 \}$$
in the interval $b\in \left(0,\tfrac{1}{2}\right)$.
If it's not possible to solve exactly, I would also appreciate to understand the scaling behavior of $a(b)$.
As a first step, even a numerical analysis would help, I simply lack the Mathematica skills do get what I want.
 A: Using the Lagrange multipliers. First we make some variable changes:
$$
\cases{
a=u^2\\
x = v^2
}
$$
so we arrive at the lagrangian
$$
L(u,v,\lambda,e) = u^2+\lambda\left(u^2v^{10}-b v^4+2v^2-1+e^2\right)
$$
and the stationary points are given by the solutions to
$$
\left\{
\begin{array}{rcl}
 u\left(\lambda v^{10}+1\right)&=&0 \\
 \lambda\left(5 u^2 v^9-2 b v^3+2v\right) &=&0\\
 u^2 v^{10}-b v^4+2 v^2-1+e^2&=&0 \\
 e \lambda&=&0  \\
\end{array}
\right.
$$
so we have the options
$$
\cases{
2 u\left(\lambda  v^{10}+1\right)=0\Rightarrow\{ u = 0\}\cup \{\lambda v^{10}+1=0\}\\
 e \lambda = 0\Rightarrow \{ \lambda = 0\}\cup\{ e = 0\}
}
$$
now multiplying $5 u^2 v^9-2 b v^3+2v$  by $v$ and the third by $5$ we have
$$
\cases{
5 u^2 v^{10}-2 b v^4+2v^2 = 0\\
5u^2 v^{10}-5b v^4+10 v^2-5+5e^2=0 
}
$$
so we follow with an easy polynomial and after some substitutions we arrive at the solution.
$$
a = f(b) = \frac{1}{625} \left(30 \left(\sqrt{16-15 b}-15\right) b^2+32 \left(40-7 \sqrt{16-15 b}\right) b+\frac{1024}{5} \left(\sqrt{16-15 b}-4\right)\right)
$$
Follows a plot showing $a = f(b)$

