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.
 A: 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.
A: 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.
A: Completing the squares
$$x^2+3x+\frac1x -\frac{15}4=\left(x-\frac12\right)^2+4x+ \frac1x -4 = \left(x-\frac12\right)^2+ \left(2\sqrt x-\frac1{\sqrt x}\right)^2\ge0$$
A: 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.
A: There is the following general statement, which we can prove by AM-GM.

Let $f$ be a polynomial of one variable $x$ with positive greatest coefficient and has unique  negative coefficient. Also, for some positive $a$ we have $$f(a)=f'(a)=0.$$ Prove that $f(x)\geq0$ for any non-negative $x$.

Your inequality follows immediately from this statement for $a=\frac{1}{2}.$
A: $$\dfrac{a\cdot\dfrac{x^2}a+3b\cdot\dfrac xb+c\cdot\dfrac1{cx}}{a+3b+c}\ge\sqrt[a+b+c]{\left(\dfrac{x^2}a\right)^a\cdot\left(\dfrac xb\right)^{3b}\cdot\left(\dfrac1{cx}\right)^c}$$
The coefficient of $x$ under radical is $$2a+3b-c$$
Let us set $2a+3b-c=0$
$$\implies\dfrac{a\cdot\dfrac{x^2}a+3b\cdot\dfrac xb+(2a+3b)\cdot\dfrac1{(2a+3b)x}}{a+3b+(2a+3b)}\ge\sqrt[a+b+(2a+3b)]{\left(\dfrac{x^2}a\right)^a\cdot\left(\dfrac xb\right)^{3b}\cdot\left(\dfrac1{(2a+3b)x}\right)^{2a+3b}}$$
$$\implies\dfrac{x^2+3x+\dfrac1x}{3(a+2b)}\ge (a^{-a}b^{-3b}(2a+3b)^{-(2a+3b)})^{1/(3(a+2b))}$$
WLOG $a=b=1$
$$\implies\dfrac{x^2+3x+\dfrac1x}{9}\ge (5^{-5})^{1/9}$$
It is sufficient to show $$\dfrac9{5^{-5/9}}\ge\dfrac{15}4\iff12\ge5^{1-5/9}\iff12^9\ge5^4$$ which is obvious
A: 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 $y=1 \Leftrightarrow x=\frac 12$.
