# 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

• You tried to what both sides? Oct 4 '12 at 8:12

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.

• Nice solution.. Oct 4 '12 at 10:05
• Good perspective, the problem becomes very simple and intuitive.Several other users on both sides of the inequality cubic, why a dead end it?Because there is little difference between what? Oct 4 '12 at 10:58

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 $\sqrt[3]60>2+\sqrt[3]7$.

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 $$8993 = 23^2\cdot 17 > 67^2\cdot 2 = 8978$$

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: $$45>6\sqrt[3]{7\times 60} \Rightarrow 45^3=91125 >6^3420=216\cdot 420=90720$$

• Proved $\displaystyle\frac{\frac{60-15}{6\sqrt[3]{7}}}{\sqrt[3]{60}}=\frac{60-15}{6 \sqrt[3]{7} \sqrt[3]{60}} >1$, does not seem to address substantive issues? Oct 4 '12 at 11:01
• You can repeat the process to make the lhs bigger and bigger, since $\left(2+\sqrt[3]{7} \right)$ reproduces. Conversely, start with $59$ instead of $60$ and it decreases... Oct 4 '12 at 11:06
• Your thinking is very special I need to think about, thank you Oct 4 '12 at 11:09
• Maybe your example is chosen in a special way... Oct 4 '12 at 11:10

There are no variables, so why not just calculate it? $$60^{1/3}>2+7^{1/3}$$ $$3.914...>3.912...$$

• ...and how do you do that without a calculator? Oct 4 '12 at 11:53
• The poster didn't say that a calculator is not allowed. Oct 4 '12 at 11:54