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Prove that $$\sin x + 2x \ge \frac{3x.(x+1)}{\pi}\quad\forall x \in \left[0,\frac{\pi}{2}\right].$$

My work: $$ 3x^2 + (3-2\pi )x - \pi \sin x \le 0 $$ $$ f(x) = 3x^2 + (3-2\pi )x -\pi \sin x $$ $$f(0)=0 $$ $$f\left({\pi\over 2 }\right) ={\pi\over 2 }\left(1 -{\pi\over 2 } \right)$$

What should I do next ?

Théophile
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Aakash Kumar
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  • Find $f'(x)$ and $f''(x)$ and hence determine if $f(x)$ is increasing/decreasing in the given interval. Use the fact that if $g(x)$ is an increasing function and $x>0$, then $g(x)>g(0)$ – Saransh Kumar Jul 13 '16 at 14:57
  • $f$ isn't increasing or decreasing in the given interval my $f$ is $$\pi\sin(x)+2\pi x-3x^2-3x$$ – Dr. Sonnhard Graubner Jul 13 '16 at 15:11
  • $f$ has a maximum in the given interval since $$f''(x)=-\pi\sin(x)-6<0$$ the first derivative is given by $$f'(x)=\pi\cos(x)+2\pi -6x-3$$ and is strictly monotonously decreasing – Dr. Sonnhard Graubner Jul 13 '16 at 15:15

2 Answers2

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Hint: We have $f''(x) = 6 + \pi\sin x$. And since $\sin x \ge -1$ for all $x$, then $f''(x) \ge 5 > 0$ for all $x$.

Now use the information at the endpoints: $f(0)=0$ and $f\left({\pi\over 2 }\right) ={\pi\over 2 }\left(1 -{\pi\over 2 } \right) < 0$.

Théophile
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The $\bf{L.H.S}$ is Concave and its graph passes through $(0,0)$ and $\displaystyle \left(\frac{\pi}{2},1+\pi\right)$.

Hence it is stay above the line Connecting These $2$ points.

And $\bf{R.H.S}$ is Convex and its graph passes through $(0,0)$ and $\displaystyle \left(\frac{\pi}{2},\frac{3}{2}+\frac{3\pi}{4}\right)$

Hence it is stay below the line connecting these $2$ points.

bcz $\displaystyle 1+\pi>\frac{3}{2}+\frac{3\pi}{4}$. So the first line is above the second line , Giving the results.

juantheron
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