## Saturday, April 25, 2015

### Hard Inequality Problem: Prove that $6<3^{\sqrt{3}}<7$

Prove that $6<3^{\sqrt{3}}<7$ from the simple and straightforward inequality where $1<\sqrt{3}<2$.

This is a particularly daunting mathematics challenge and you could spend days or even a week trying to prove the inequality, with no fruitful result.

Notice that if we exponentiate the given inequality $1<\sqrt{3}<2$ with base 3, we get:

$3^1<3^{\sqrt{3}}<3^2$

$3<3^{\sqrt{3}}<9$ (Compare it with the targeted inequality $6<3^{\sqrt{3}}<7$)

The lower and upper bound that we could get from the given inequality is far too low and high for the targeted inequality.

In other words, we need to think for another credible plan of attack, and we need a much narrow gap between the lower and upper bound for $\sqrt{3}$.

But for any of the given inequality, what we are allowed to do are:

1. to perform addition and/or subtraction of a quantity to both sides, or

2. to divide and/or multiply by some positive quantity without changing the inequality signs (or with some negative quantity by changing the inequality signs at the same time),

3. to square or to take both sides to $n$ power,

3. or to take logarithm to both sides or

4. to exponentiate both sides with the known base.

With our ultimate goal (that is to narrow down the gap) in mind, we tried the following:

First attempt (we subtract all $3$ parts of the given inequality by $\dfrac{1}{2})$:

$1<\sqrt{3}<2$

$1-\dfrac{1}{2}<\sqrt{3}-\dfrac{1}{2}<2-\dfrac{1}{2}$

$\dfrac{1}{2}<\sqrt{3}-\dfrac{1}{2}<\dfrac{3}{2}$

We then square each side to get:

$\left(\dfrac{1}{2}\right)^2<\left(\sqrt{3}-\dfrac{1}{2}\right)^2<\left(\dfrac{3}{2}\right)^2$

$\dfrac{1}{4}<3-\sqrt{3}+\dfrac{1}{4}<\dfrac{3}{2}$

$\dfrac{1}{4}<\dfrac{13}{4}-\sqrt{3}<\dfrac{3}{2}$

$\dfrac{1}{4}-\dfrac{13}{4}<\dfrac{13}{4}-\sqrt{3}-\dfrac{13}{4}<\dfrac{3}{2}-\dfrac{13}{4}$

$-3<-\sqrt{3}<-\dfrac{7}{4}$

$\dfrac{7}{4}<\sqrt{3}<3$

$\large\therefore 3^{\frac{7}{4}}<3^{\sqrt{3}}<3^3$

Even though we don't know offhand if $\large3^{\frac{7}{4}}>6$ or otherwise, we can still tell immediately the upper bound is too large, and therefore this method by first subtracting all three parts by $\dfrac{1}{2})$ doesn't help much. But this did reveal that we need to subtract something bigger than $\dfrac{1}{2})$.

Do we stop and put this problem under the carpet, and tell everyone that we tried, but we failed with a hurdle and we don't think this problem can be solved using any elementary method.

Or, do you keep plugging away until you savor the taste of success? I am fairly certain that everyone want to be the latter, because we are born to explore, and leave no stones unturned.

Second attempt (we subtract all $3$ parts of the given inequality by $1$:

$1<\sqrt{3}<2$

$1-1<\sqrt{3}-1<2-1$

$0<\sqrt{3}-1<1$

We then square each side to get:

$0^2<(\sqrt{3}-1)^2<1^2$

$0<3-2\sqrt{3}+1<1$

$0<4-2\sqrt{3}+1<1$

$-4<-2\sqrt{3}<-3$

$3<2\sqrt{3}<4$

$\dfrac{3}{2}<\sqrt{3}<2$

$\large\therefore 3^{\frac{3}{2}}<3^{\sqrt{3}}<3^2=9$

This second attempt is futile attempt as well because we have to prove $3^{\sqrt{3}}<7$.

This one time, we will try to subtract not $\dfrac{1}{2}$ nor $1$ but $\dfrac{3}{2}$ and see where that leads us.

Third attempt (we subtract all $3$ parts of the given inequality by $\dfrac{3}{2})$:

$1<\sqrt{3}<2$

$1-\dfrac{3}{2}<\sqrt{3}-\dfrac{3}{2}<2-\dfrac{3}{2}$

$-\dfrac{1}{2}<\sqrt{3}-\dfrac{3}{2}<\dfrac{1}{2}$

We then square each side to get:

$\left(-\dfrac{1}{2}\right)^2<\left(\sqrt{3}-\dfrac{3}{2}\right)^2<\left(\dfrac{1}{2}\right)^2$

$0<\left(\sqrt{3}-\dfrac{3}{2}\right)^2<\left(\dfrac{1}{2}\right)^2$

$0<3-3\sqrt{3}+\dfrac{9}{4}<\dfrac{1}{4}$

$0<\dfrac{21}{4}-3\sqrt{3}<\dfrac{1}{4}$

$-\dfrac{21}{4}<-3\sqrt{3}<-5$

$5<3\sqrt{3}<\dfrac{21}{4}$

$\dfrac{5}{3}<\sqrt{3}<\dfrac{7}{4}$

$\large\therefore 3^{\frac{5}{3}}<3^{\sqrt{3}}<3^{\frac{7}{4}}$

If we can prove $\large 6<3^{\frac{5}{3}}$ and $\large 3^{\frac{7}{4}}<7$, then we are done.

Yes! We can see the light at the end of the tunnel at this point in time!

It's not difficult to see that

a.) $\large 6<3^{\frac{5}{3}}$ holds since $\large 6^3=216<3^5=243$ and

b.) $\large 3^{\frac{7}{4}}<7$ holds since $\large 3^7=2187<7^4=2401$

and hence we have proved that $6<3^{\sqrt{3}}<7$ from $1<\sqrt{3}<2$.

Yes, we just put up a titanic fight with this problem but we did learn something invaluable today, that is, perseverance will yield rich reward!