<s>[INST] Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.
 [/INST] Here is a verification approach by factorization:
We want to show
$$x^2+3x+\frac1x  \ge \frac{15}4$$
We know that $x > 0$,
Hence it is equivalent to showing:
$$4x^3 + 12x^2 + 4 \ge 15x$$
$$4x^3+12x^2-15x+4 \ge 0$$
Observe that $x=-4$ is a root.
$$(x+4)(4x^2-4x+1) \ge 0$$
$$(x+4)(2x-1)^2 \ge 0$$
Since $x>0$, clearly, $x+4>0$ and $(2x-1)^2 \ge 0$, hence the product must be nonnegative.
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