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Consider the double inequality : $$6<3^\sqrt3 < 7$$ Using $only$ the elementary properties of exponents and inequalities (NO calculator, computer, table of logarithms, or estimate of √3 may be used), prove that the first inequality implies the second.

SOURCE : "Inequalities proposed in Crux Mathematicorum" (Page Number 14; Question Number 627)

I have no idea how to solve this problem. I tried taking logarithms, but that did not help. I strongly suspect that some inequality has to be used. I tried AM-GM on a few set of terms, but that just made it messier.

Wolfram Aplha gives the value of ${3^{\sqrt {3}}}$ as $\approx 6.7049918538258$.

Can anyone give me a hint on how to solve this problem ?

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    So are we for instance not allowed to use that $1.7<\sqrt3<1.75$? – 2017-01-16
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    Without estimates of any kind, we just can't move on. See, we might just as well use $6<4^\sqrt4$, but that most definitely doesn't imply $4^\sqrt4<7$. – 2017-01-16
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    Since there is now way to estimate the value of $3^{\sqrt{3}}$, one has to treat it as unknown. And $6 – 2017-01-16
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    @Laray I have stated the problem as it was given. It is from the proposed questions for CRUX(link : http://imomath.com/pcpdf/f1/f41.pdf), and moreover it was proposed $WITH$ a solution... – 2017-01-16
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    @Laray : $1<\sqrt{2}$ implies $\sqrt{2}<2$, indeed multiply both sides of the first inequality with $\sqrt{2}$ and the second follows. Why something similar could not work for OP's problem, I don't see. – 2017-01-16
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    @Raskolnikov Because you can't rid of $\sqrt 3$ in exponent by given operations unless you jump through quite a number of hoops. Basically, you're given $3^{\sqrt 3 - 1} > 2$ (which is also hard to prove normally) "for free", this lets you to remove degrees of $2$ from inequalities. – 2017-01-16
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    @Abstraction Did you know that $(3^{\sqrt3})^{\sqrt3}=27$? I don't know whether this leads to a solution, but I can't say immediately that it doesn't. – 2017-01-16
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    @Arthur How does that solve the problem ???? – 2017-01-16
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    It doesn't, but it was close. What I thought was that raising to the power of $\sqrt3$ gives $$6^{\sqrt3}<27\\ 2^{\sqrt3}3^{\sqrt3}<28$$ and then the idea was to divide by $4$, but I saw that $\frac{2^{\sqrt3}}{4}$ is on the wrong side of $1$ for me to conclude anything. – 2017-01-16
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    It seems very unlikely that any proof of $6 < 3^{\sqrt 3} \vdash 3^{\sqrt 3} <7$ would actually use the premise in a necessary way, since the conclusion is a tighter bound than the premise ($3^{\sqrt 3}$ is closer to $7$ than to $6$). – 2017-01-16
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    @DanielV: I find this argument a little weak. $f(3^{\sqrt3})$ is closer to $f(6)$ than $f(7)$ for a suitable monotonous $f$, such as $f(x)=\exp(2(x-3^\sqrt3))$. – 2017-01-16
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    @YvesDaoust Even simpler, and perhaps more relevant: If you raise all the terms to some power, the distance between the two larger numbers will grow faster than the distance between the smaller numbers. Specifically, [WA](http://www.wolframalpha.com/input/?i=solve+(3%5Esqrt(3))%5Ex+-+6%5Ex+%3D+7%5Ex+-+(3%5Esqrt(3))%5Ex) says it happens around $13.3$ in our case. – 2017-01-17
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    You use Wolfram Alpha... but the problem includes ***NO*** calculator, computer, etc. – 2018-02-18

2 Answers 2

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Claim: $7 > 3^{\sqrt 3}$. Otherwise $$7^\sqrt 3 \le 27 < 28 = 7*4 \\ 7^{\sqrt 3 - 1} < 4 \\ 48 < 7^2 < 4^{\sqrt 3 + 1} \text{(since }(7^{\sqrt 3 - 1})^{\sqrt 3 + 1} = 7^{(\sqrt 3 - 1)(\sqrt 3 + 1)} = 7^{(\sqrt 3)^2 - 1^2} = 7^2{)} \\ 3 < 4^{\sqrt 3 - 1} \: \text{(divide by 16)}\\ 3^{\sqrt 3 + 1} < 4^2 = 16 < 18 \\ 3^\sqrt 3 < 6$$ ...and we have a contradiction with a given statement.

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    I worked out what you did, but I think this answer would be more clear if you explained what algebraic operations you did at each step. Not all proofs can be sone without words. – 2017-01-16
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    I think this can be turned upside-down (plus reversing all the inequalities) and be a direct proof without resorting to contradiction. But otherwise a nice find. – 2017-01-17
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$$6=(6^3)^{\frac{1}{3}}=216^{\frac{1}{3}}<243^{\frac{1}{3}}=(3^5)^{\frac{1}{3}}=3^{\frac{5}{3}}<3^{\sqrt{3}},$$ becouse $$\left(\frac{5}{3}\right)^2=\frac{25}{9}<3=\left(\sqrt{3}\right)^2.$$ Since $$\left(\sqrt{3}\right)^2=3=\frac{48}{16}<\frac{49}{16},$$ then $$\sqrt{3}<\frac{7}{4}.$$

Therefore $$3^{\sqrt{3}}<3^{\frac{7}{4}}=(3^7)^\frac{1}{4}=2187^{\frac{1}{4}}<2401^{\frac{1}{4}}=(7^4)^{\frac{1}{4}}=7.$$

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    Nice answer.+1 @afalnik... – 2017-01-16
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    Proved the inequalities are true, but did you actually use the first inequality to prove the second? – 2017-01-16