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If $x^Ty\leq1$ for all $y$ with $\lVert{y}\rVert_2=1$, then $\lVert{x}\rVert_2\leq{1}$. $x,y\in R^n$

I have tried to prove it by using the definition of vector inner product: $$x^Ty=\lVert{x}\rVert_2\lVert{y}\rVert_2cos\theta$$ But this definition is generally used in a space whose dimension is equal to or less than 3, so a better proof is required here.

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  • $\begingroup$ Are you guys satisfied with the improvement of my question? : ) $\endgroup$ – BioCoder Aug 23 '15 at 11:48
  • $\begingroup$ Why is this question still "on hold" ? I have improved it as required. $\endgroup$ – BioCoder Aug 24 '15 at 8:00
  • $\begingroup$ The body of the Question should be as self-contained as possible, not relying on the title to bear the burden of posing the problem. What you wrote as the "definition" of vector inner product is really a geometric interpretation, relating the scalar value of the inner product to an angle of incidence between the (nonzero) vectors $x,y$. An algebraic definition of inner product can be found in your textbook or on this Wikipedia page. $\endgroup$ – hardmath Aug 28 '15 at 3:12
  • $\begingroup$ Hardmath,thanks for your comment $\endgroup$ – BioCoder Sep 5 '15 at 8:03
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Hint Set $$y =\frac{x}{\| x\|_2}$$

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  • $\begingroup$ Cool ! I like your solution, which is so elegant! $\endgroup$ – BioCoder Aug 23 '15 at 3:15

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