# Minimum value of $\frac{(x+8)^3}{x}$ without calculus, and weighted AM-GM inequality

So I was trying to find the minimum value of $$f(x) = \frac{(x+8)^3}{x}, x>0$$ without calculus but something weird happended when applying weighted AM-GM inequality.

First of all, of course $$f(x) = x^2 + 24x + 512 x^{-1} + 192$$. So I decided to find the minimum by applying the weighted AM-GM inequality: $$1(x^2) + 1(24x) + 3 (\frac{512}{3} x^{-1}) + 1(192) \geq 6 (x^2 \times 24 x \times 512^3 3^{-3} x^{-3} \times 192)^{1/6} = 128 × 3^{5/6}.$$ Which is about 319.25. To my surprise, however, this is not the minimum value of the function (the minimum is actually f(x) = 432). Why did this happen? I usually do find correct values to the function minima when I apply the AM-GM inequality. What's different this time?

• You have a bound that would be attainable if $x^2=24x=512x^{-1}/3=192$, and under the assumption that $x\geq0$. To make those inequalities true, one would need $x=\frac{512}{192\cdot 3}$ (from the last equation), but this doesn't satisfy the first equation. Nov 20, 2019 at 21:57

You don't get the minimum because equality can not hold in your application of the AM-GM inequality: $$x^2 = 24 x = \frac{512}{3} x^{-1} = 192$$ is not possible.

If you start with $$x+8 = x + 4 + 4 \ge 3 \sqrt[3]{x \cdot 4 \cdot 4}$$ then you'll get $$\frac{(x+8)^3}{x} \ge 27 \cdot 16 = 432$$ which is sharp because equality holds for $$x=4$$.

• I see, that makes sense. Thanks! Nov 20, 2019 at 22:08

As an alternative, $$k\in\mathbb{R}^+$$ belongs to the range of $$f(x)=\frac{(x+8)^3}{x}$$ over $$\mathbb{R}^+$$ iff $$(x+8)^3-kx=0$$ has three real solutions, i.e. iff the discriminant of $$(x+8)^3-kx$$ is $$\geq 0$$, i.e. iff $$4k^3-1728k^2\geq 0$$, i.e. iff $$k\geq 432$$.

• I don't really see why "iff $(x + 8)^3 - kx = 0$ has three real solutions". Couldn't we just have a single real solution? Nov 20, 2019 at 22:12
• @JoãoVítorG.Lima: if it has a single real solution, such solution is negative, since $(x+8)^3/x$ is convex over $\mathbb{R}^+$ and $\lim_{x\to 0^+}=\lim_{x\to +\infty}=+\infty$. Nov 20, 2019 at 22:12

The AM-GM inequality will give you the minimum of the function if and only if there exists some specific value $$x$$ such that all coefficient in the inequality are equal. In your case iff $$\exists x, \ x^2=24x=\frac{512}{3x}=192$$ Which is not the case.

I have never used the inequality to find the minimum of a function, so I don't know why/when it should work.