2
$\begingroup$

$\left(\frac{\sqrt{2017}-1}{45-\sqrt{3}}\right)^{2016}+\left(\frac{\sqrt{2017}+1}{45+\sqrt{3}}\right)^{2016}>2$

Can someone help me ?

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    Have you tried rationalizing the denominators? – 2017-01-14
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    Ahm yes i tried that and it will look like $ \left(\frac{\left(\sqrt{2017}-\:1\right)\left(45\:+\sqrt{3}\right)}{2022}\right)^{2016}+\left(\frac{\left(\sqrt{2017}+1\right)\left(45\:-\sqrt{3}\right)}{2022}\right)^{2016}>\:2$ – 2017-01-15
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    Okay, that gives me an idea. But it suggests you have a typo someplace. Note that the $\pm 1$ in the numerators is under the radical in the Question, but outside the radicals in your Comment. – 2017-01-15
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    How is looks in the comment is correct , in the question i accidentally put 2017-1 under the radical – 2017-01-15
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    I've edited the Question to reflect your last Comment. Please review to make sure it's correct. – 2017-01-15
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    Yes , now it is correct thanks , do you know what could i do next , after rationalizing ? – 2017-01-15
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    It's a bit of a strange question, in that the first term by itself is already greater than $2$, as you can check with a calculator. I'm tempted to think this is an arithmetic-geometric mean problem that got slightly garbled. If you have the original source of the problem, I would check that the $3$'s in the denominator are supposed to have square roots on them. – 2017-01-15
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    This problem can be found in a gazette from my country , in the gazette 3 has a square root on it – 2017-01-15

1 Answers 1

-1

$$\left(\frac{\sqrt{2017}-1}{45-\sqrt{3}}\right)^{2016}+\left(\frac{\sqrt{2017}+1}{45+\sqrt{3}}\right)^{2016}>\left(\frac{\sqrt{2017}-1}{45-\sqrt{3}}\right)^{2016}>1.01^{2016}>1+2016\cdot0.01>2$$ Done!

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    @John L. What do you think about my solution? – 2017-05-10