<s>[INST] Prove that $6<3^\sqrt3$ implies $3^\sqrt3 < 7$ 
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 ?
 [/INST] $$6=(6^3)^{\frac{1}{3}}=216^{\frac{1}{3}}<243^{\frac{1}{3}}=(3^5)^{\frac{1}{3}}=3^{\frac{5}{3}}<3^{\sqrt{3}},$$ 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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