Prove $\sqrt[3]{60}>2+\sqrt[3]{7}$ Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$
I try to both sides of the cubic equation, but it is quite complicated
 A: For $a,b>0$ and $a\neq b$, $(a^{3}+b^{3})-(a^{2}b+ab^{2})=(a-b)(a^{2}-b^{2})>0$, then $(a+b)^{3}=(a^{3}+b^{3})+3(a^{2}b+ab^{2})<4(a^{3}+b^{3})$, set $a=2=\sqrt[3]8$ and $b=\sqrt[3]7$, we can obtain that $(2+\sqrt[3]7)^{3}<60$, so $\sqrt[3]60>2+\sqrt[3]7$.
A: Well, I don't see any alternative way..
$$60\overset{?}> 8+6\sqrt[3]{7^2}+12\sqrt[3]7+7 $$
$$45\overset{?}> 6\sqrt[3]{7^2}+12\sqrt[3]7 $$
$$15\overset{?}> 2\sqrt[3]{7^2}+4\sqrt[3]7 $$
Well, we could raise it to cubic, but that's really not nice.
What about considering the roots of $2x^2+4x-15 =2(x+1)^2-17$, and finally comparing if $\sqrt[3]7$ is between its roots..
$$\sqrt{\frac{17}2} -1 \overset{?}> \sqrt[3]7 $$
A bit nicer perhaps.. taking cubes:
$$\frac{17}{2}\sqrt{\frac{17}{2}}-3\cdot\frac{17}2+3\cdot\sqrt{\frac{17}2}-1 \overset{?}>7
$$
$$23\sqrt{\frac{17}2} \overset{?}> 16+3\cdot 17 = 67$$
and finally this leads to
$$8993 = 23^2\cdot 17 > 67^2\cdot 2 = 8978 $$
A: One can avoid brute-force approach just using the concavity of $f(x) = \sqrt[3]{x}$. Any strictly concave function satisfies the following relation:
$$ f \left(\frac{x+y}{2}\right) > \frac{f(x) + f(y)}{2} $$
After setting $f(x) = \sqrt[3]{x}$, $x = 8$ and $y = 7$ one obtains exactly the same inequality as in the question.
A: When you cube it, you get:
$$
 60 > \left(2+\sqrt[3]{7} \right)^3 =15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
Continue to get
$$
 \frac{60-15}{6\sqrt[3]{7}} > \left(2+\sqrt[3]{7} \right).
$$
Now cube again to get:
$$
 \left(\frac{60-15}{6\sqrt[3]{7}}\right)^3 > 15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
The rhs is now larger than at the beginning, because $\displaystyle\frac{\frac{60-15}{6\sqrt[3]{7}}}{\sqrt[3]{60}}=\frac{60-15}{6\sqrt[3]{7}\sqrt[3]{60}} >1$. 
To show this we use:
$$
45>6\sqrt[3]{7\times 60} \Rightarrow 45^3=91125 >6^3420=216\cdot 420=90720
$$
A: There are no variables, so why not just calculate it?
$$60^{1/3}>2+7^{1/3}$$
$$3.914...>3.912...$$
