<s>[INST] Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots. 
Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.

For an equation to have nonnegative roots, it mustn't only have negative roots.
Let $y = x^2 - x + 4$ $(y > 0)$, we have that $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = mx^2 + (2m - 1)xy + my^2$$
For the equation $$mx^2 + (2m - 1)xy + my^2 = 0$$ to have only negative roots, it must be satisfied that $\left\{ \begin{align} [(2m - 1)y]^2 - 4m^2y^2 \ge 0\\ (1 - 2m)y < 0\\ my^2 > 0 \end{align} \right.$ $\implies \left\{ \begin{align} (1 - 4m)y^2 \ge 0\\ 1 - 2m < 0\\ m > 0 \end{align} \right.$ $\implies \left\{ \begin{align} 1 - 4m \ge 0\\ m > \dfrac{1}{2}\\ m > 0 \end{align} \right.$ $\iff \left\{ \begin{align} m \le \frac{1}{4}\\ m > \dfrac{1}{2} \end{align} \right. \implies m \in \varnothing$.
Thus for $\forall m \in \mathbb R$, the equation $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
Is the above solution correct? And should my attempt was inaccurate and yours is (hopefully) helpful, please post an answer.
 [/INST] Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $x^2$
$$0=m(x^2+16/x^2)+x+4/x+8m-1=m(x+4/x)^2+x+4/x-1$$
Now as $x\ge0,$ $$\dfrac{x+4/x}2\ge\sqrt{x\cdot4/x}=?$$
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