<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] The AM-GM inequality would not be any better than derivatives at finding your minimum. But if you know (or suspect) that the minimum is at $x = \frac12$, then we can use it to get a proof.
First: at $x = \frac12$, the three terms are equal to $\frac14$, $\frac32$, and $2$. If we want to use a weighted AM-GM inequality to prove that $x^2 + 3x + \frac1x \ge \frac{15}{4}$, this has to be the equality case. So, we rewrite $x^2 + 3x + \frac1x$ as
$$
   \frac1{15} (15 x^2) + \frac25 \left(\frac{15}{2}x\right) + \frac{8}{15} \left(\frac{15}{8x}\right)
$$
because the three terms $15x^2$, $\frac{15}{2}x$, and $\frac{15}{8x}$ are equal at $x=\frac12$. (We get the coefficients $\frac1{15}$, $\frac25$, and $\frac{8}{15}$ by normalizing $\frac14$, $\frac32$, and $2$ to add to $1$.)
Now, we apply weighted AM-GM:
$$
   \frac1{15} (15 x^2) + \frac25 \left(\frac{15}{2}x\right) + \frac{8}{15} \left(\frac{15}{8x}\right) \ge (15 x^2)^{1/15} \left(\frac{15}{2}x\right)^{2/5} \left(\frac{15}{8x}\right)^{8/15}.
$$
The right-hand side simplifies to $\frac{15}{4}$, as all the powers of $x$ cancel, and we get the lower bound we wanted.

The general procedure at work here is called geometric programming. Usually, we do not have a candidate solution, and we will try arbitrary weights $\delta_1, \delta_2, \delta_3 \ge 0$ with $\delta_1 + \delta_2 + \delta_3 = 1$ (to apply AM-GM) and $2\delta_1 + \delta_2 - \delta_3 = 0$ (so that the powers of $x$ cancel). Then we would get a lower bound on $x^2 +3x + \frac1x$ depending on these weights, and we would optimize our choice of weights to get the best lower bound possible.
In this example, this technique doesn't work any better than setting the derivative equal to $0$. It really shines in multivariable examples.
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