Compare $A$ and $B$ with: $$A = \sqrt{2017} + \sqrt{2019} + \sqrt{2023}$$ $$B = \sqrt{2018} + \sqrt{2020} + \sqrt{2021}$$ I tried to prove $A^4 < B^4$ but it's too hard to do that.
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1Quick beginner guide for asking a well-received question + please avoid "no clue" questions. – Anne Bauval Jul 12 '23 at 15:51
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1Nguyen, could you write your attempts and efforts in your post? – Angelo Jul 12 '23 at 16:48
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Well its enough to prove $$A < B$$ since taking power to a positive number is operation that does not change sign. Easy to prove seems $$A^2 < B^2$$ just by doing all of that products.
Quick guess without calculator would be also to use Taylor series $$\sqrt{2017 + x} \approx \sqrt{2017} + \frac{x}{2\sqrt{2017}} - \frac{1}{8}\frac{x^2}{2017^{3/2}} $$ so only third term makes difference and you effectively want to compare $x^2$ $$2^2+6^2 > 1^2 + 3^2 + 4^2$$ so that $$40 > 26$$ seems to hold.
VojtaK
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In the same spirit : replace $2017$ by $x$ and use Taylor series for infintely large $x$ to get $$A-B=-\frac{7}{4 x^{3/2}}\left( 1-\frac{33}{7 x}+O\left(\frac{1}{x^2}\right)\right)$$ – Claude Leibovici Jul 12 '23 at 12:32
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Thank you but I had done it already. I use the subtraction of A and B; it seems easier than yours. – Nguyen Huy Gia Bao Jul 12 '23 at 12:42
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1This question goes against our quality standard policy. Instead of posting an answer, you should encourage the person who posted the question to improve it. – Anne Bauval Jul 12 '23 at 15:51