<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] A bit late but I thought worth mentioning it:
Consider the equivalent inequality
$$4x^2+12x+ \frac 4x = (2x)^2 + 6(2x)+\frac 8{2x}\geq 15$$
Setting $y=2x$, you get
\begin{eqnarray*} y^2+6y+\frac 8y
& = & y^2 - 2y +8\left(y+\frac 1y\right)\\
& = & (y-1)^2-1+8\left(y+\frac 1y\right) \\
& \geq & 0-1+8\cdot 2 = 15 
\end{eqnarray*}
with equality if and only if $y=1 \Leftrightarrow x=\frac 12$.
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