<s>[INST] Show that the following inequality holds $\sqrt[3]{2}+\sqrt2>2$ Show that the following inequality holds $$\sqrt[3]{2}+\sqrt2>2$$
I have just studied cubic roots and we know that if $a<b$ then $\sqrt[3]{a}<\sqrt[3]{b}$ $(a,b\in\mathbb{R}).$ Here though we also have the square root $\sqrt[2]2$. How are we supposed to approach the problem? Thank you. The approximate values are $$\sqrt[3]{2}\approx1.26;\sqrt2\approx1.41$$
So using @rtybase suggestion we have to show $\sqrt[3]2+\sqrt2>2$. We know that $\sqrt{2}>1$. Can we subtract the second inequality from the first? Is subtracting inequalities even a thing? Then we will get $\sqrt[3]2>1=\sqrt[3]{1}$ which is true.
 [/INST] We have:
$$\begin{array}{l}
\sqrt[3]{2} > 2 - \sqrt{2} \iff 2 > (2 - \sqrt{2})^3\iff \\
\iff 2 > 8 - 3\cdot 2^2\cdot \sqrt{2}+3\cdot 2\cdot (\sqrt{2})^2- 2\sqrt{2}=20-14\sqrt{2}\iff \\ \iff14\sqrt{2} > 18\iff 7\sqrt{2} > 9\iff 98 > 81  \end{array}
$$,
which is true.
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