<s>[INST] 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?
 [/INST] 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.
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