8

For any non-zero real number $x_1$, $x_2$,$$\left|x_1\sin{\frac{1}{x_1}}-x_2\sin{\frac{1}{x_2}}\right|\leq2\sqrt{|x_1-x_2|}.$$ I tried the following, $$\begin{array}{rl}\left|x_2\sin\frac{1}{x_2}-x_1\sin\frac{1}{x_1}\right|&\leq \left|x_2\sin\frac{1}{x_2}-x_1\sin\frac{1}{x_2}\right|+\left|x_1\sin\frac{1}{x_2}-x_1\sin\frac{1}{x_1}\right|\\ &\leq |x_2-x_1|+|x_1|\left|\frac{1}{x_2}-\frac{1}{x_1}\right|\\ &=\frac{1+|x_2|}{|x_2|}|x_2-x_1|\\ \end{array}$$ But I don't know how to go on...

Hilbert1994
  • 1,159
  • 2
    Does this answer your question? https://math.stackexchange.com/questions/523297/how-to-prove-this-inequality-bigx-sin-frac1x-y-sin-frac1y-big2 – Math Lover Jan 16 '21 at 08:53
  • @Math Lover well thank you,I have a look at it. It seems that there is no strict proof.btw,how do you find these posts? – Hilbert1994 Jan 16 '21 at 09:03
  • Nothing fancy, I just noticed it on the right, under section Linked – Math Lover Jan 16 '21 at 09:05
  • @Math Lover I feel these answers are not very beautiful – Hilbert1994 Jan 16 '21 at 09:07
  • 1
    Yes you may be right but also consider this - the question had a bounty on it which would usually attract higher attention to it. So may be there is no better solution to it. – Math Lover Jan 16 '21 at 09:14
  • @MathLover The bounty was given by Community so it need not have a bounty set up by asker – Albus Dumbledore Jan 16 '21 at 09:39
  • @AlbusDumbledore you are missing the point. It does not matter who awarded the bounty. The bounty is to get more attention to the question. – Math Lover Jan 16 '21 at 10:52

0 Answers0