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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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    $\begingroup$ 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 $4^\sqrt4<7$. $\endgroup$ – Ivan Neretin Jan 16 '17 at 9:16
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    $\begingroup$ Since there is now way to estimate the value of $3^{\sqrt{3}}$, one has to treat it as unknown. And $6<x$ can never imply $x<7$. $\endgroup$ – Laray Jan 16 '17 at 9:16
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    $\begingroup$ @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. $\endgroup$ – Raskolnikov Jan 16 '17 at 9:32
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    $\begingroup$ @Arthur How does that solve the problem ???? $\endgroup$ – user399078 Jan 16 '17 at 9:49
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    $\begingroup$ @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 $f(x)=\exp(2(x-3^\sqrt3))$. $\endgroup$ – Yves Daoust Jan 16 '17 at 11:14
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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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    $\begingroup$ 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. $\endgroup$ – Oscar Lanzi Jan 16 '17 at 11:34
  • $\begingroup$ 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. $\endgroup$ – Arthur Jan 17 '17 at 14:32
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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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    $\begingroup$ Nice answer.+1 @afalnik... $\endgroup$ – user401699 Jan 16 '17 at 10:07
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    $\begingroup$ Proved the inequalities are true, but did you actually use the first inequality to prove the second? $\endgroup$ – David Jan 16 '17 at 10:10

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