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Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$

I try to both sides of the cubic equation, but it is quite complicated

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You tried to what both sides? – Chris Eagle Oct 4 '12 at 8:12
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Oct 14 '12 at 12:50
up vote 14 down vote accepted

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.

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Nice solution.. – Berci 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? – geometryscience 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 $\sqrt[3]60>2+\sqrt[3]7$.

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

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

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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? – geometryscience 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... – draks ... Oct 4 '12 at 11:06
Your thinking is very special I need to think about, thank you – geometryscience Oct 4 '12 at 11:09
Maybe your example is chosen in a special way... – draks ... Oct 4 '12 at 11:10

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

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

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