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What is the general method for finding such inequalities? I have some more problems of this kind in the text I am using.

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By the AM-GM Inequality, $$\frac{2+1+1+1+1+1+1+1+1+1}{10} \geq \sqrt[10] 2$$ $$\frac{11}{10} \geq \sqrt[10] 2 \Rightarrow 1.2 \geq \sqrt[10] 2$$ The AM-GM inequality sometimes works when you have an n-th root but it depends on the problem.

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Using the binomial theorem, $$1.2^{10}=\left(1+\frac 15\right)^{10}\gt 1+\binom{10}{1}\frac 15=3\gt 2\Rightarrow 1.2\gt \sqrt[10]2.$$

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  • $\begingroup$ I think part of the beauty of it at first was in not showing that the binomial theorem was being used. This made people think for a moment. :) $\endgroup$ – Daniel W. Farlow Apr 6 '15 at 23:19
  • $\begingroup$ @MagicMan: I thought so. But later, I began to think adding the link might be better:) (partly because I got several downvotes...) $\endgroup$ – mathlove Apr 6 '15 at 23:22
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    $\begingroup$ Fair point. And maybe it helped OP--which is, after all, the point. Kudos. $\endgroup$ – Daniel W. Farlow Apr 6 '15 at 23:24
  • $\begingroup$ Using the Bernoulli inequality is easier (but equivalent, in this case). $\endgroup$ – egreg Apr 6 '15 at 23:32
  • $\begingroup$ @egreg: I agree with you. Didn't notice it. $\endgroup$ – mathlove Apr 6 '15 at 23:41
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There is no general method. There have been numerous books written on just inequalities. For example, see Inequalities: Theorems, Techniques and Selected Problems. However, oftentimes, simple inequalities like yours may be approached by applying some subtle manipulations, much as mathlove's excellent answer has done. Consider the following example.

Example: Compare $6^{1/4}$ and $4^{1/3}$ without using a calculator. Is one greater than the other?

Answer. Observe that $$ 6^{\frac{1}{4}}=6^{\frac{3}{12}}=(6^3)^{\frac{1}{12}}=216^{\frac{1}{12}} $$ and $$ 4^{\frac{1}{3}}=4^{\frac{4}{12}}=(4^4)^{\frac{1}{12}}=256^{\frac{1}{12}}. $$ Since $256$ and $216$ are both being raised to the same power, and $256>216$, we can clearly see that $4^{1/3} > 6^{1/4}$ since $$ 4^{\frac{1}{3}} = 256^{\frac{1}{12}} > 216^{\frac{1}{12}} = 6^{\frac{1}{4}}. $$


Knowing that it helps to make such manipulations comes with time and practice. Notice that it would otherwise be quite difficult to compare the values in the example as $4^{1/3}\approx 1.5874$ and $6^{1/4}\approx 1.5651$.

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Try raising $1.2$ to higher and higher powers: $$ (1.2)^2=1.44 > 1.4, $$ $$ (1.2)^4 > (1.4)^2 = 1.96 > 1.9, $$ $$ (1.2)^5 > (1.2)(1.9)=2.28 > 2. $$ So already $(1.2)^5 > 2$, hence $1.2 > \sqrt[5]{2} > \sqrt[10]{2}$.

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$\sqrt[10]2\lt\sqrt[8]2$. $\sqrt[8]2$ is easy to compute by $\sqrt{\sqrt{\sqrt2}}$, which is certainly $\lt 1.2$.

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Think of an investment that returns $20\%$ per year (and let me know if you can find one...). Simple interest would return $100\%$ -- i.e. double -- in five years, hence quadruple in ten. Compound interest, which is what $1.2^{10}$ represents, necessarily does better, so $1.2$ is quite a bit larger than $\sqrt[10]2$.

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    $\begingroup$ I don't understand the downvote. This is another way of taking the tenth power and saying $1.2^{10} \gt 1+10(1^9)0.2=3 \gt 2$ $\endgroup$ – Ross Millikan Apr 7 '15 at 2:47

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