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.

If we know equality when $x=\frac{1}{2}$ then find AM-GM solution very easy.
For $x=\frac 12,$ then
$$x^2 = \color{blue}{\frac{1}{4}}, \quad 3x = \frac{3}{2} = \color{red}{6} \cdot \color{blue}{\frac{1}{4}}, \quad \frac 1x = 2 = \color{red}{8} \cdot \color{blue}{\frac{1}{4}}.$$
Now, using the AM-GM inequality, we have
$$x^2+3x+\frac1x = x^2 + 6 \cdot \frac x 2+8 \cdot \frac{1}{8x} \geqslant 15\sqrt[15]{x^2\left(\frac x 2\right)^6\left(\frac{1}{8x}\right)^8}=\frac{15}{4}.$$
Done.
