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.
Let $\,p(x)=ax^5-bx^2+2x-1\,$ with $\,a \gt 0\,$ and $\,0 \lt b \lt \frac{1}{2}\,$.
$\,p(0) = - 1 \lt 0\,$ and $\,p(1) = a - b + 1 \gt 0\,$ since $\,a \gt 0 \gt b-1\,$.
$\,p'(x)=5ax^4-2bx+2 \gt 0\,$ for $\,0 \le x \le 1\,$ since $\,-2bx+2 \ge-2b+2 \gt 0\,$.
By the rule of signs, $\,p'(x)\,$ has no negative roots, and $\,0\,$ or $\,2\,$ positive ones.
If $\,p'(x)\,$ has no positive roots, then $\,p'(x) \gt 0\,$ on $\,\mathbb R^+\,$, so $\,p(x)\,$ is strictly increasing and, given $\,p(1) \gt 0\,$, it follows that $\,p(x)\,$ has no roots $\,x \ge 1\,$.
If $\,p'(x)\,$ has $\,2\,$ positive roots, they must both be $\,\ge 1\,$ since $\,p'(x) \gt 0\,$ on $\,[0,1]\,$.
Therefore, a necessary condition for $\,p(x)\,$ to have a root $\,\ge 1\,$ is for $\,p'(x)\,$ to have $\,2\,$ real roots, which will then both be $\,x_{1,2}'\gt 1\,$.
$\,p''(x)=20ax^3-2b\,$ has a unique real root $\,x_1'' = \sqrt[3]{\frac{b}{10a}} \gt 0\,$, which corresponds to a minimum of $\,p'(x)\,$. The sufficient condition for $\,p'(x)\,$ to have $\,2\,$ real roots is that the value at the minimum point be negative or zero $\,p'(x'') \le 0\,$:
$$ \require{cancel} 0 \ge p'(x'') = 5\cancel{a}\frac{b}{10\cancel{a}}\sqrt[3]{\frac{b}{10a}} - 2b\sqrt[3]{\frac{b}{10a}}+2 = -\frac{3b}{2}\sqrt[3]{\frac{b}{10a}} + 2 $$ $$ \iff\quad\quad a \le a_0 = \frac{27 b^4}{640} \tag{1} $$
Piecing together the findings above, when $\,(1)\,$ is satisifed, $\,p(x)\,$ will have a root in $\,(0,1)\,$, then will increase to a maximum at $x_1' \gt 1$, decrease to a minimum at $\,x_2' \ge x_1'\,$, then increase to $\,+\infty\,$. The necessary and sufficient condition for a root $\,\gt 1\,$ to exist is that $\,p(x_2') \le 0\,$, and in that case there will be two such roots on either side of $\,x_2'\,$.
When $\,a \to 0\,$ the equation reduces to a quadratic with two real roots $\,\gt 1\,$. As $\,a\,$ increases, the dominant term in $\,p(x)\,$ is $\,ax^5\,$ when $\,x \gt 1\,$, so the "∿" shaped graph of $\,p(x)\,$ gets "pulled up". The two real roots get closer and closer to one another, until at some point they become equal then vanish altogether. The point where that happens is when $\,p(x)\,$ has a double root i.e. its discriminant becomes $\,0\,$:
$$ \begin{align} 0 = \Delta &= a^2 (3125 a^2 + 4500 a b^2 - 12800 a b + 8192 a + 108 b^5 - 108 b^4) \\ &= a^2 \big(3125 a^2 + 4(1125 b^2 - 3200 b + 2048) a + 108 b^4(b-1)\big) = a^2\,q(a) \end{align} $$
The part in parenthesis $\,q(a)\,$ is a quadratic in $\,a\,$, and it can be verified that its discriminant $\,\delta \gt 0\,$ for $\,0 \lt b \lt \frac{1}{2}\,$, so the quadratic has two real roots of opposite signs $\,a_1 \lt 0 \lt a_2\,$ since their product is negative. Then the condition for $\,p(x)\,$ to have real roots is (courtesy WA):
$$ a \le a_2 = - \frac{2}{3125}\left(2048 - 3200 b + 1125 b^2 - (32 - 5 b)(16 - 15 b)\sqrt{16 - 15 b}\right) \tag{2} $$
Taking the minimum between $\,(1),(2)\,$ gives in the end $\,a \le a_{max} = \min\left(a_0, a_2\right)\,$.
For verification, consider for example $\,b=\frac{1}{4}\,$, then $\,a_0 =$ $27/163840\,$, $\,a_2=$ $3/50000\,$, so $\,a_{max} = 3/50000\,$. Indeed, for $\,a=a_{max}\,$ the equation becomes:
$$ \frac{3}{50000} x^5 - \frac{1}{4} x^2 + 2 x - 1 = \frac{1}{50000}(x - 10)^2 (3 x^3 + 60 x^2 + 900 x - 500) $$
When $\,a = a_{max}\,$ the equation has the double root $\,x=10\,$, and for $\,0 \lt a \lt a_{max}\,$ it will have two distinct real roots $\,\gt 1\,$, in addition to the third real root in $\,(0,1)\,$ which always exists.
[ EDIT ] $\;$ It can be verified that $\,q(a_0) \gt 0\,$ when $\,0 \lt b \lt \frac{1}{2}\,$, which means $\,a_0 \not\in [a_1,a_2]\,$, and given that $\,a_0\,$ is positive it follows that $\,a_0 \gt a_2\,$, so in the end $\,a_{max}=\min(a_0,a_2)=a_2\,$.