Sign of $P(Δ)$ with $P(x)=ax²+bx+c$ and $Δ=b²-4ac$

Let $P(x)=ax^2+bx+c$ be a polynomial with $a≥0$ and $b≥\frac{1}{8a}$, and $Δ=b^2-4ac$. Show that $P(Δ)≥0$.

My answer: We have $Δ=b^2-4ac$. If$Δ≤0$, then $P(Δ)≥0$. If $Δ>0$, there are two cases:

If $c≥0$, $P(Δ)=aΔ^2+bΔ+c≥0$. If $c<0,Δ=b^2-4ac$ then $Δ>b^2$ and $Δ>-4ac$, then $$P(Δ)=aΔ^2+bΔ+c>a×(-4ac)^2+b^3+c=16a^3c^2+c+b^3$$ is of degree two of variable $c$. We have $Δ_1=1-4×16a^3b^3=1-(4ab)^3$, which is negative if $b≥\frac{1}{4a}$ or we have $b≥\frac{1}{8a}$.

I want the correct solution of the exercise.

Solution 1:

Completing the square differently, this is non-negative when $\,a \gt 0\,$, $\,\Delta \ge 0\,$, $\,2b - \dfrac{1}{4a} \ge 0\,$:

$$ \begin{align} \frac{1}{a} P(\Delta) &= \Delta^2+\frac{b}{a}\Delta \color{red}{-2 \frac{b}{a}\Delta + \frac{b^2}{4a^2} + 2 \frac{b}{a}\Delta - \frac{b^2}{4a^2}}+\frac{c}{a} \\ &= \Delta^2 - 2 \frac{b}{2a}\Delta + \frac{b^2}{4a^2}+\frac{2b}{a}\Delta-\frac{b^2-4ac}{4a^2} \\ &= \left(\Delta - \frac{b}{2a}\right)^2 + \left(2 b - \frac{1}{4a}\right)\frac{\Delta}{a} \end{align} $$