text: # good trignometric inequality
  
  • Apr 7th 2013, 10:16 PM
  earthboy
  good trignometric inequality
  prove that $\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$
  • Apr 7th 2013, 10:30 PM
  earthboy
  Re: good trignometric inequality
  I started by trying to show the expression as the sum of squares:
  $\displaystyle x^2\sin x + x \cos x + x^2 + \frac{1}{2}$
  $\displaystyle = x^2\sin x + x \cos x + x^{2}(\sin^{2} x + \cos^{2} x)+\frac{1}{4}+ \frac{1}{4}$
  $\displaystyle =(x^{2} \cos^{2} x + x \cos x + \frac{1}{4})+x^{2} \sin^{2} x + x^{2} \sin x + \frac{1}{4}$
  $\displaystyle =(x \cos x + \frac{1}{2})^2 + x^2 \sin^2 x + x^2 \sin x + \frac{1}{4}$
  
  any suggestion on how to go about from here......???
  
  Advanced thanks for any help or suggestions....(Happy)
  • Apr 7th 2013, 10:48 PM
  Gusbob
  Re: good trignometric inequality
  Quote:
  
  Originally Posted by earthboy
  prove that $\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$
  
  Wow this is a fun inequality. This is what I did, it is similar to yours.
  
  $\displaystyle 0\leq\frac{(x\sin(x)+x)^2}{2}=\frac{1}{2}(x^2\sin^ 2(x) + 2x^2\sin(x)+x^2)$
  
  $\displaystyle 0\leq\frac{(x\cos(x)+1)^2}{2}=\frac{1}{2}(x^2\cos^ 2(x) + 2x\cos(x)+1)$
  
  $\displaystyle 0\leq \frac{1}{2}(x^2(\sin^2(x)+\cos^2(x))+2(x^2\sin(x)+ x\cos(x))+x^2+1)$
  
  $\displaystyle 0\leq x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}$
  
  I'm not sure about if the strict inequality case is true. If it is, the adjustment would be minor (just show that if one of the expressions is zero if the other one is strictly positive).
  • Apr 8th 2013, 04:42 AM
  topsquark
  Re: good trignometric inequality
  Quote:
  
  Originally Posted by earthboy
  prove that $\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$
  
  I did it pretty much term by term.
  
  Note that
  $\displaystyle x^2~sin(x) + x^2 = x^2(sin(x) + 1)$
  As sin(x) has a minimum of -1, this expression is always 0 or positive.
  
  Note that
  $\displaystyle x~cos(x) \geq 0$ since both x and cos(x) are odd functions.
  
  And of course 1/2 is always positive.
  
  -Dan
  • Apr 8th 2013, 10:42 PM
  earthboy
  Re: good trignometric inequality
  (Ninja)great proof! Gusbob!
  
  Quote:
  
  Originally Posted by topsquark
  Note that
  $\displaystyle x~cos(x) \geq 0$ since both x and cos(x) are odd functions.
  
  -Dan
  
  Thanks Dan!(Nod), but can you or anybody please explain how this part is true ???
  As $\displaystyle x$ is any real number, what about if $\displaystyle x=120$ , then $\displaystyle x \cos x = -60$ which is $\displaystyle \leq 0$.
  
  • Apr 15th 2013, 12:23 AM
  ohiosuba
  Re: good trignometric inequality
  great proof! Gusbob!
  
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  • Apr 15th 2013, 07:15 AM
  topsquark
  Re: good trignometric inequality
  Quote:
  
  Originally Posted by earthboy
  (Ninja)great proof! Gusbob!
  
  Thanks Dan!(Nod), but can you or anybody please explain how this part is true ???
  As $\displaystyle x$ is any real number, what about if $\displaystyle x=120$ , then $\displaystyle x \cos x = -60$ which is $\displaystyle \leq 0$.
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