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Inequality using Calculus

jacks

Well-known member
Apr 5, 2012
226
Prove that $\displaystyle \sin x+2x \geq \frac{3x(x+1)}{\pi}\forall x\in \left[0,\frac{\pi}{2}\right]$
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
I'm tempted to "use calculus" to produce the Taylor Series for f(x) = sin(x) and then observe that:

[tex]\sin(x) < x - \frac{x^{3}}{6}[/tex]
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
I think I would approach it this way: minimize the function
$$f(x)=\sin(x)+2x-\frac{3x(x+1)}{\pi}$$
on the interval $[0,\pi/2]$ using the standard calculus techniques. There's one critical point near $0.88$, and then you have the endpoints. The left endpoint, I think, will end up being the smallest point on the graph. The second derivative might come in handy.
 

awkward

Member
Feb 18, 2012
36
Here is a slightly different approach.

Using f(x) as defined in post #3, show that \( f''(x) < 0 \) for all x, so f is concave. By computation, show that \( f(0) = 0 \) and \( f(\pi / 2) > 0 \). This is enough to conclude that \( f(x) > 0 \) on \( [0 , \pi/2] \).