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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?

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    $\begingroup$ 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. $\endgroup$ Nov 20 '19 at 21:57
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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 $x=4$.

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  • $\begingroup$ I see, that makes sense. Thanks! $\endgroup$ Nov 20 '19 at 22:08
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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 $k\geq 432$.

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  • $\begingroup$ 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? $\endgroup$ Nov 20 '19 at 22:12
  • $\begingroup$ @JoãoVítorG.Lima: if it has a single real solution, such solution is negative, since $(x+8)^3/x$ is convex over $\mathbb{R}^+$ and $\lim_{x\to 0^+}=\lim_{x\to +\infty}=+\infty$. $\endgroup$ Nov 20 '19 at 22:12
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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.

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