# How do I prove $\lVert{x}\rVert_2\leq{1}$

## 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.

• Are you guys satisfied with the improvement of my question? : ) – BioCoder Aug 23 '15 at 11:48
• Why is this question still "on hold" ? I have improved it as required. – BioCoder Aug 24 '15 at 8:00
• 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. – hardmath Aug 28 '15 at 3:12
• Hardmath,thanks for your comment – BioCoder Sep 5 '15 at 8:03

Hint Set $$y =\frac{x}{\| x\|_2}$$