If the roots of $ax^3 - x^2 + bx - 1$ are real and positive, prove $b \geq \sqrt{3}$ Let 
$$p(x) = ax^3 - x^2 + bx - 1$$
such that $a, b \in \mathbb{R}$ and the roots of $p(x)$ are real and positive.
Prove that 
$$b \geq \sqrt{3}$$
Let $\alpha, \beta, \gamma \in \mathbb{R^{+}}$ be the roots of $p(x)$
$$\alpha + \beta + \gamma = \sigma_1 = \frac{1}{a}$$
$$\alpha\beta + \beta\gamma + \gamma\alpha = \sigma_2 = \frac{b}{a}$$
$$\alpha\beta\gamma = \sigma_3 = \frac{1}{a}$$
Consider the inequality:
$$\alpha^2 + \beta^2 + \gamma^2 \geq \alpha\beta + \beta\gamma + \gamma\alpha$$
Since the inequality is cyclic, we may assume $\alpha \geq \beta \geq \gamma$. Now, the above is clearly true by the rearrangement inequality.
$$\implies \sigma_1^2 - 2\sigma_2 \geq \sigma_2$$
$$\implies \sigma_1^2 \geq 3\sigma_2$$
$$\implies \frac{1}{a^2} \geq 3\frac{b}{a}$$
$$\implies 1 \geq 3ab \tag{1}$$
Now, from the AM-GM inequality:
$$\alpha + \beta + \gamma \geq 3\sqrt[3]{\alpha\beta\gamma}$$
Since $\alpha + \beta + \gamma = \alpha\beta\gamma$,
$$\alpha^3\beta^3\gamma^3 \geq 27\alpha\beta\gamma$$
$$\implies \alpha\beta\gamma = \frac{1}{a} \geq 3\sqrt{3}$$
$$\implies \frac{1}{3\sqrt{3}} \geq a$$
Putting this in $(1)$, we get:
$$3ab \leq \frac{b}{\sqrt{3}}$$
We know that $3ab \leq 1$, but we can't say that $\frac{b}{\sqrt{3}}\leq 1$ as well. Did I go wrong somewhere? Or is there some way to solve it from here?

Everything you have derived is true, but it's not enough to finish the problem.  Here is one way to do it:
We have $\sigma_2^2 = \alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 + 2(\alpha+\beta+\gamma)(\alpha\beta\gamma) = \alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 + 2\sigma_1\sigma_3$.  Now we have that $\displaystyle \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} \geq \frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} +\frac{1}{\alpha\gamma}$, and multiplying this by $(\alpha\beta\gamma)^2$ we get $\alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 \geq (\alpha+\beta+\gamma)(\alpha\beta\gamma) = \sigma_1\sigma_3$. Hence we get $\displaystyle \frac{b^2}{a^2} = \sigma_2^2 \geq 3\sigma_1\sigma_3 = \frac{3}{a^2}$, or $b^2 \geq 3$.  Therefore $b\geq\sqrt{3}$ as required.
