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Find the sum of all positive integers $n,$ where the inequality $$\sqrt{a + \sqrt{b + \sqrt{c}}} \ge \sqrt[n]{abc}$$holds for all nonnegative real numbers $a,$ $b,$ and $c.$

What I have tried:

I have already tried squaring both sides but that got me nowhere.

I also have tried the AM - GM inequality and this is what I have:

$$\frac {a + b + c + (n - 3) \cdot 1} {n} \ge \sqrt [n] {abc}.$$

I'm mainly looking for hints, but answers are welcome.

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    $\begingroup$ what I see is different ranges. First, if $a=1, b=1, c > 1$ and $c$ large, the left hand side is essentially $\sqrt[8] c.$ Slightly larger, call it $\delta + \sqrt[8] c$ where $\delta > 0 $ is thought of as small. For what $n$ is that larger than $ \sqrt[n] c \; \; ?$ Next, various test cases with $abc < 1$ which is where you get upper bounds on $n$ $\endgroup$
    – Will Jagy
    Commented Aug 2, 2023 at 2:01
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    $\begingroup$ Do you mean both sides to have n terms [you only have 3 factors] $\endgroup$
    – coffeemath
    Commented Aug 2, 2023 at 2:02

4 Answers 4

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The first thing I would do is introduce variables that I can think of as "being in the same units" (e.g., meters). Thus, let $a = u$, $b = v^2$, and $c = w^4$. Then we may write the inequality as$$ u + \sqrt{v^2 + w^2} \ge x^{14/n}, \;\text{where}\; x = (u v^2 w^4)^{1/7}. $$ Here we are thinking of $u$, $v$, $w$, and $x$ as being quantities in meters. If $n = 14$ then both sides of the inequality are in the same units. But otherwise the two sides are in different units, which seems strange. How can we exploit this? E.g., if $n < 14$ what happens when $u$, $v$, and $w$ are very large? What happens when $n > 14$? Finally, what happens when $n = 14$?

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  • $\begingroup$ Correct me if I'm wrong, but the way I'm interpreting the "units", is the degree/power of the variable? $\endgroup$ Commented Aug 2, 2023 at 3:17
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    $\begingroup$ Yes, rather than doing applied math, we can keep things pure. The expressions $u + \sqrt{v^2 + w^2}$ and $(u v^2 w^4)^{1/7}$ are homogeneous functions with degree of homogeneity = $1$, whereas $x^{14/n}$ has degree $14/n$. $\endgroup$
    – Jim Ferry
    Commented Aug 2, 2023 at 3:29
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    $\begingroup$ @WIlliamRLi the term is "dimensional analysis" and you would see mathematicians refer to dimensional analysis as rescaling, that is for what $$\begin{cases}a\mapsto\lambda a \\ b\mapsto \lambda^sb \\ c\mapsto \lambda^t c\end{cases}$$ is the scale factor $\lambda$ homogeneous i.e. leaves the term invariant because some power of $\lambda$ can be factored out equally? $\endgroup$ Commented Aug 2, 2023 at 3:31
  • $\begingroup$ Two other hints for the $n = 14$ case: power mean inequality and weighted A.M.-G.M. $\endgroup$
    – Jim Ferry
    Commented Aug 2, 2023 at 3:35
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Firstly $b+\sqrt{c}=b+\sqrt{c}/2+\sqrt{c}/2\ge 3\sqrt[3]{bc/4}$.

Then $a + \sqrt{b + \sqrt{c}}\ge a+\sqrt{3}\sqrt[6]{bc/4}=a+\frac{\sqrt{3}\sqrt[6]{bc/4}}{6}+\cdots+\frac{\sqrt{3}\sqrt[6]{bc/4}}{6}\ge 7\sqrt[7]{Cabc}$.

Here $C$ is a constant which equals $\frac{1}{4}(\sqrt{3}/6)^6$. Then just need to check $7\sqrt[7]{C}\ge 1$ to see why $n=14$ works. And you can show that this is the minimized $a + \sqrt{b + \sqrt{c}}$ when you fix $abc$, which proves other $n$ fails.

Sorry for the messy notion, hope you can grasp the idea.

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  • $\begingroup$ I finally understand this, thank you. $\endgroup$ Commented Aug 2, 2023 at 14:26
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After beautiful Peter Wu's and Jim Ferry's solutions,

I want to add a bit of simpler proof for $n=14.$

Let $\sqrt{a+\sqrt{b+\sqrt{c}}}=x.$

Thus, $$x\geq\sqrt{a},$$ $$x\geq\sqrt[4]{b}$$ and $$x\geq\sqrt[8]c.$$ Thus, $$x^{2+4+8}\geq abc$$ and $$\sqrt{a+\sqrt{b+\sqrt{c}}}\geq\sqrt[14]{abc}.$$

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    $\begingroup$ I understand after the clear explanation of Jim Ferry that the only thing left to prove is that $u+\sqrt{v^2+w^2}> (uv^2w^4)^{1/7}.$ It seems pretty obvious by taking $u=1$ without loss of generality, and $ v=r\cos \theta,\ w=r\sin\theta$ ; therefore $$1+r>r^{6/7}> r^{6/7}(\cos \theta)^{2/7}(\sin \theta)^{4/7.}$$ $\endgroup$ Commented Aug 2, 2023 at 8:20
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    $\begingroup$ I find Michael's proof for the $n=14$ case particularly elegant. Both this and Gerard's are nicer than the cookbook approach had in mind: $(u+2v+4w)/7 \ge (uv^2w^4)^{1/7}$ and $\sqrt{(v^2+w^2)/2} \ge (v+w)/2$. $\endgroup$
    – Jim Ferry
    Commented Aug 2, 2023 at 11:50
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    $\begingroup$ @William R Li I think my contribution to the solution is not worth accepting. Give please accept to Peter Wu or to Jim Ferry. $\endgroup$ Commented Aug 2, 2023 at 14:10
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    $\begingroup$ But I love your solution so much @MichaelRozenberg $\endgroup$ Commented Aug 2, 2023 at 14:49
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    $\begingroup$ Your solution helps me understand the idea of solving inequality problems like this. $\endgroup$ Commented Aug 2, 2023 at 15:31
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$\textbf{Hint:}$ Use a change of variables such that

$$\begin{cases}u=a\\ v^2=b+\sqrt{c}\\ w=b-\sqrt{c}\end{cases}\implies u+v\geq\left(\frac{u(v^4-w^2)(v^2-w)}{8}\right)^{\frac{2}{n}}$$

This means we can reduce this to the problem, finding which $n$ does the following inequality hold?

$$u+v\geq\left(\frac{u(v^4-w^2)(v^2-w)}{8}\right)^{\frac{2}{n}}$$

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  • $\begingroup$ You are such a genius thanks bro $\endgroup$ Commented Aug 2, 2023 at 2:10
  • $\begingroup$ Wait, theres 1 problem: (v^4 - w^2) = (v^2 + w) (v^2 - w) = (2b)(2sqrt(c)) NOT 2b2c $\endgroup$ Commented Aug 2, 2023 at 2:23
  • $\begingroup$ @WilliamRLi you're right, I have edited the problem. The third degree of freedom remains $\endgroup$ Commented Aug 2, 2023 at 2:28
  • $\begingroup$ I'm still having trouble continuing though..... $\endgroup$ Commented Aug 2, 2023 at 3:21

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