prove that$\displaystyle x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}>0$ for any real number $\displaystyle x$

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- Apr 7th 2013, 10:16 PMearthboygood 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 PMearthboyRe: 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 PMGusbobRe: good trignometric inequality
Wow this is a fun inequality. This is what I did, it is similar to yours.

$\displaystyle 0\leq\frac{(x\sin(x)+x)^2}{2}=\frac{1}{2}(x^2\sin^ 2(x) + 2x^2\sin(x)+x^2)$

$\displaystyle 0\leq\frac{(x\cos(x)+1)^2}{2}=\frac{1}{2}(x^2\cos^ 2(x) + 2x\cos(x)+1)$

Adding the two gives

$\displaystyle 0\leq \frac{1}{2}(x^2(\sin^2(x)+\cos^2(x))+2(x^2\sin(x)+ x\cos(x))+x^2+1)$

$\displaystyle 0\leq x^{2} \sin x + x \cos x + x^{2} + \frac{1}{2}$

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 AMtopsquarkRe: good trignometric inequality
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.

Add them all up and you get your inequality.

-Dan - Apr 8th 2013, 10:42 PMearthboyRe: good trignometric inequality
(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 $\displaystyle \leq 0$.

anyways..advanced thanks as usual!!!(Pizza) - Apr 15th 2013, 12:23 AMohiosubaRe: good trignometric inequality
great proof! Gusbob!

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Purchase the most fantastic series and The Bible DVD canada with high qaulity for free home delivery! - Apr 15th 2013, 07:15 AMtopsquarkRe: good trignometric inequality