Question

How would one show that
$\Large\textrm{ \sqrt[3]{3}+\sqrt{2}\ \textgreater \ \sqrt[3]{9}\ \textgreater \ \sqrt[4]{13} }$

No calculator, no approximation allowed.

Answer 1

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We need to prove,

$\rm\sqrt[3]{3}+\sqrt{2}>\sqrt[3]{9}>\sqrt[4]{13}$

$\;$

So, what we need to show is,

$\begin{cases} &\rm\sqrt[3]{3}+\sqrt{2}>\sqrt[3]{9} \\\\ &\rm \sqrt[3]{9}>\sqrt[4]{13} \end{cases}$

$\;$

$\textbf{- Showing the First Inequation}$

Isolating square roots and cube roots,

$\rm\sqrt{2}>\sqrt[3]{9}-\sqrt[3]{3}$

$\;$

Let the value $\rm\sqrt[3]{9}-\sqrt[3]{3}$ be $\rm A$.

$\;$

We know,

$\boxed{\rm{(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)}}$

$\rm{A^{3}}$

$\;$

$\rm=(\sqrt[3]{9})^{6}-(\sqrt[3]{3})^{3}-3\cdot(\sqrt[3]{9\cdot3})^{2}\cdot(\sqrt[3]{9}-\sqrt[3]{3})$

$\;$

$\rm =(3^{2}-3)-3\sqrt[3]{3^{3}}\cdot(\sqrt[3]{9}-\sqrt[3]{3})$

$\;$

$\rm =(3^{2}-3)-9(\sqrt[3]{9}-\sqrt[3]{3})$

$\;$$\;$

$\rm =6-9A$

$\;$

$\rm\therefore A=\sqrt[3]{9}-\sqrt[3]{3}\ \Longrightarrow\ A^{3}=6-9A$

$\;$

We know,

$\boxed{\rm a^{n}>b^{n}>1\iff a>b>1\ (n>1)}}$

$\rm2^{3}<9<3^{3}$

$\;$

$\rm\therefore2<\sqrt[3]{9}<3$

$\;$

We know,

$\boxed{\rm a^{n}>b^{n}>1\iff a>b>1\ (n>1)}}$

$\rm1^{3}<3<2^{3}$

$\;$

$\rm\therefore1<\sqrt[3]{3}<2$

$\;$

By taking the difference of the inequations,

$0<\sqrt[3]{9}-\sqrt[3]{3}<2$

$\;$

$\rm\therefore0[tex]\;$

By cubing,

$\rm1[tex]\;$

$\rm 1[tex]\;$

$\rm 1<6-9A<8$

$\;$

$\rm\dfrac{2}{9}[tex]\;$

$\dfrac{2}{9}<\sqrt[3]{9}-\sqrt[3]{3}<\dfrac{5}{9}$

$\;$

By squaring,

$\rm1^{2}<2<2^{2}\iff1<\sqrt{2}<2$

$\;$

Then we obtain,

$\rm\dfrac{2}{9}<\sqrt[3]{9}-\sqrt[3]{3}<\dfrac{5}{9}<1<\sqrt{2}<2$

$\;$

$\therefore\sqrt[3]{9}-\sqrt[3]{3}<\sqrt{2}$

$\;$

Hence shown that,

$\therefore\boxed{\rm{\sqrt[3]{3}+\sqrt{2}>\sqrt[3]{9}}}$

$\;$

$\textbf{- Showing the Second Inequation}$

We know,

$\boxed{\rm{a^{n}>b^{n}>1\iff a>b>1}}$

$\rm(\sqrt[3]{9})^{12}=9^{4}=6561,\ (\sqrt[4]{13})^{12}=13^{3}=2197$

$\;$

Hence shown that,

$\rm\therefore\boxed{\rm{\sqrt[3]{9}>\sqrt[4]{13}}}$

$\;$

$\textbf{- Result}$

$\therefore\boxed{\rm{\sqrt[3]{3}+\sqrt{2}>\sqrt[3]{9}>\sqrt[4]{13}}}$

$\;$

$\Large\textrm{Learn More}$

$\textbf{- Addition in Inequalities}$

$\boxed{\rm a_{1}[tex]\;$

$\textbf{- Subtraction in Inequalities}$

$\boxed{\rm a_{1}[tex]\;$

To find the product or division, consider the signs first.

Let's consider positive numbers only.

$\textbf{- Multiplication in Inequalities}$

$\boxed{\rm a_{1}[tex]\;$

$\textbf{- Division in Inequalities}$

[tex]\boxed{\rm a_{1}