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To find the minimum value of $x+\dfrac{1}{x}$, $(x>0)$ we can use the AM-GM inequality to say that $$\dfrac{x+\dfrac{1}{x}}{2}\geq \sqrt{x\cdot \dfrac{1}{x}}$$ or $$x+\dfrac{1}{x}\geq 2$$ The minimum value is when $x=\dfrac{1}{x}$ i.e. at $x=1$

If we take another function $4x^{3}+\dfrac{1}{x}$ then $$\dfrac{4x^{3}+\dfrac{1}{x}}{2}\geq \sqrt{4x^{3}\cdot \dfrac{1}{x}}$$ or $$4x^{3}+\dfrac{1}{x}\geq 4x$$ Here the minimum value should be $2\sqrt{2}$ at $4x^{3}=\dfrac{1}{x}$ i.e. at $x=\dfrac{1}{\sqrt{2}}$ but the actual minimum value is at $x=\dfrac{1}{\sqrt[4] {12}}$.
This is just a random example but I've noticed that if the variables don't cancel out on the GM side then the wrong minimum value is obtained. What am I doing wrong?
(Note: I am aware that there are better methods to find the minimum value. I am just wondering about this specific method.)

Sagewizard88
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2 Answers2

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Plotting the curves $4x^3+1/x$ and $4x$ might help you understand the issue. The AM-GM inequality always holds. The blue curve is above the red one, except in one point. But that point is not the minimum of the blue curve.

enter image description here

Andrei
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    Sorry @JohnOmielan. It's just a typo in my text. The function plotted is indeed $4x^3+1/x$ – Andrei Dec 12 '22 at 20:45
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Use AM-GM like this:

$$4x^3+\frac{1}{x}=4x^3+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}\geq4(4x^3\cdot \frac{1}{3x}\cdot \frac{1}{3x}\cdot \frac{1}{3x})^\frac{1}{4}=4(\frac{4}{27})^\frac{1}{4}$$

Itoz Darbien
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