# Proving An Inequality

#### anemone

##### MHB POTW Director
Staff member
Deduce from the simple estimate that if $$\displaystyle 1<\sqrt{3}<2$$, then $$\displaystyle 6<3^{\sqrt{3}}<7$$.

Hi members of the forum,

This problem says the resulting inequality may be deduced from the simple estimate, but I was unable to do so; could anyone shed some light on how to deduce the intended result?

#### Nehushtan

##### New member
Re: Proving an inequality

$$\displaystyle 1\,<\,\sqrt3\,<\,2$$​

$$\displaystyle \Rightarrow\ -\frac12\,<\,\sqrt3-\frac32\,<\,\frac12$$

$$\displaystyle \Rightarrow\ 0<\,\left(\sqrt3-\frac32\right)^2\,<\,\frac14$$

$$\displaystyle \Rightarrow\ 0<\,\frac{21}4-3\sqrt3\,<\,\frac14$$

$$\displaystyle \Rightarrow\ \frac53<\,\sqrt3\,<\,\frac74$$

$$\displaystyle \Rightarrow\ 3^{5/3}<\,3^{\sqrt3}\,<\,3^{7/4}$$

Note that $$\displaystyle 6=216^{1/3}<243^{1/3}=3^{5/3}$$ and $$\displaystyle 3^{7/4}=2187^{1/4}<2401^{1/4}=7$$.

#### anemone

##### MHB POTW Director
Staff member
Re: Proving an inequality

$$\displaystyle 1\,<\,\sqrt3\,<\,2$$​

$$\displaystyle \Rightarrow\ -\frac12\,<\,\sqrt3-\frac32\,<\,\frac12$$

$$\displaystyle \Rightarrow\ 0<\,\left(\sqrt3-\frac32\right)^2\,<\,\frac14$$

$$\displaystyle \Rightarrow\ 0<\,\frac{21}4-3\sqrt3\,<\,\frac14$$

$$\displaystyle \Rightarrow\ \frac53<\,\sqrt3\,<\,\frac74$$

$$\displaystyle \Rightarrow\ 3^{5/3}<\,3^{\sqrt3}\,<\,3^{7/4}$$

Note that $$\displaystyle 6=216^{1/3}<243^{1/3}=3^{5/3}$$ and $$\displaystyle 3^{7/4}=2187^{1/4}<2401^{1/4}=7$$.

Hi Nehushtan, thanks to your simple explanation because it is now very clear to me! I appreciate it! 