14
$\begingroup$

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 ?

$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 5
    $\begingroup$ @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
  • 1
    $\begingroup$ @Arthur How does that solve the problem ???? $\endgroup$ – Nirbhay Jan 16 '17 at 9:49
  • 2
    $\begingroup$ @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
15
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
8
$\begingroup$

$$6=(6^3)^{\frac{1}{3}}=216^{\frac{1}{3}}<243^{\frac{1}{3}}=(3^5)^{\frac{1}{3}}=3^{\frac{5}{3}}<3^{\sqrt{3}},$$ 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 $$3^{\sqrt{3}}<3^{\frac{7}{4}}=(3^7)^\frac{1}{4}=2187^{\frac{1}{4}}<2401^{\frac{1}{4}}=(7^4)^{\frac{1}{4}}=7.$$

$\endgroup$
  • 1
    $\begingroup$ Nice answer.+1 @afalnik... $\endgroup$ – user401699 Jan 16 '17 at 10:07
  • 5
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.