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AM GM inequality states that for real positive numbers x,y

$$x + y \geq 2(xy)^{1/2}$$

You would get the least value for x = y

We can also write this as,

$$\frac x3 + \frac x3 + \frac x3 + y \geq 4\left((x^3)\cdot(y)\cdot\left(\frac{1}{27}\right)\right)^\frac14$$

You would get the least value for $x = 3y$

Why is this contradiction arising?

Also for a given set of variables being added together, how can we find the most accurate inequality using AM GM?

insipidintegrator
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Shreyaan
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  • I am having a tough time seeing where you pulled the $5$th line from. – copper.hat Sep 08 '22 at 16:58
  • I guess he is assuming $a + b + c + d \ge 4(abcd)^{1/4}$, and that a min occurs when $a=b=c=d$ – DanielV Sep 08 '22 at 17:01
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    It's not that you get the "least value" when x=y or x=3y, it's that the left hand side equals the right hand side at these points, but in your two equations, you have different right hand sides. – Aaron Sep 08 '22 at 17:09
  • I would rather spend time answering than unraveling questions. – copper.hat Sep 08 '22 at 17:10
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    On a side note, I have seen lots of posts like this where people express "doubt" about standard results, or think they have found flaws in them, and this is a really poor mindset for learning. The question should not be "is everybody else wrong," but rather "what am I misunderstanding?" To paraphrase Shakespeare, the fault is not in our stars, nor our theorems, but in ourselves. – Aaron Sep 08 '22 at 18:45
  • @Aaron I agree and disagree. In my view, it's -1 point for being too quick to suspect that a theorem is wrong, but +10 points for generally being skeptical! "Stress-testing" a theorem to see if it seems to lead to conclusions that don't make any sense is a good way to understand what the theorem actually does or doesn't say. – Misha Lavrov Sep 08 '22 at 19:29

2 Answers2

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Let $x,y \ge 0$.

It is true that $x + y \ge 2 (xy)^{1/2}$, and that equality holds when $x=y$.

It is also true that $x + y = \frac13x + \frac13x + \frac13 x + y \ge 4((\frac13x)^3 y)^{1/4} = \frac{4}{3^{3/4}} (x^3y)^{1/4}$, and that equality holds when $\frac13x = y$.

These do not contradict each other, and neither of them is necessarily a minimum value of $x+y$.

In some applications of $x + y \ge 2 (xy)^{1/2}$, you assume that the product $xy$ is constant. In that case and in that case only, you can say that $x+y$ is minimized when $x=y = \sqrt{xy}$.

Similarly, if the product $x^3y$ is held constant, we can use the second version of AM-GM to conclude that $x+y$ is minimized when $\frac13x = y = (\frac x3)^{3/4} y^{1/4} = \frac1{3^{3/4}} \cdot (x^3y)^{1/4}$.

These still do not contradict each other: with different conditions on $x$ and $y$, $x+y$ is minimized at different points.

Misha Lavrov
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  • Hi, I’m curious, why does the RHS have to be a constant for minima? Maybe a stupid question – insipidintegrator Sep 08 '22 at 17:21
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    If the RHS is not a constant, then you have two points $(x,y)$ of interest: the point where LHS=RHS, and the point where RHS is as small as possible. These are not the same in general, so we get a gap in the best inequality we write down. (And in these specific examples, the lower bound on the RHS will be $0$: if we fix $x^3y$, we can make $xy$ as small as we want. So the best inequality we write down will be $x+y > 0$, which is very bad.) – Misha Lavrov Sep 08 '22 at 17:24
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I'm not quite sure what you want to say in your question, but I would like to mention some points.

We consider $x$ and $y$ are non-negative.

$$x+y\ge 2\sqrt {xy}$$

That is correct. We say that the equality occurs iff (or if and only if) when $x=y$. We doesn't say $\min \{x+y\}=2\sqrt {xy}$. The argument is simple: Because, $xy$ is not a constant.

$$\frac x3+\frac x3+\frac x3+y≥4\sqrt [4]{\frac {x^3y}{27}}$$

The same can be said for the second inequality. We say that the equality occurs iff when $x=3y$. We doesn't say $\min \{x+y\}=4\sqrt [4]{\frac {x^3y}{27}}$. Because, $\frac {x^3y}{27}$ is not a constant.

In other words, the inequalities $$x+y≥ 2\sqrt {xy}$$ or $$x+y\ge 4\sqrt [4]{\frac {x^3y}{27}}$$ are both valid and true.

Which inequality to use will depend on what is intended.

Finally, consider the following exact example:

Let, $x\ge 0$ then what is the minimum value of $\dfrac 2x +x^2$ ?

If you apply the AM-GM inequality directly, you will get

$$\frac 2x +x^2\ge 2\sqrt {2x}$$

This inequality is always true and equality occurs iff $x=\sqrt [3]{2}$, however the right-hand side (RHS) is not a constant.

Now, let's apply the AM-GM inequality in a different way:

$$ \begin{aligned}\frac 2x+x^2&=\frac 1x+\frac 1x +x^2\\ &\ge 3\sqrt[3]{\frac 1{x^2}\times x^2}\\ &=3.\end{aligned} $$

You can observe that the result we want to achieve is exactly this and equality occurs iff when $x=1$.

nonstudent
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