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Discussion Response Post to: inqualitiy

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Prakhar Banga (1204)

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By the inequality, I suppose you mean justify the equality.

Now lets consider sin x/x.

Differentiating it, we get (x cos x - sin x)/x^2.

Now we know tan x>x, or x cos x< sin x, or x cos x - sin x<0.

So the derivative of sin x/x is also less than 0 in (0,pi/2]

Therefore, sin x/x decreases in this interval and its minimum value is at x=pi/2, which is sin pi/2 /(pi/2) = 2/pi.

So sin x/x >=2/pi in this interval (0, pi/2]

As pi/2>1, 3-pi/2<2, or (3-pi/2)/pi<2/pi

As sin x/x >= 2/pi and 2/pi>(3-pi/2)/pi, so sin x/x>(3-pi/2)/pi and sin x >(3-pi/2)x/pi ------ equation 1.

Also, x<=pi/2

So 3x^2<=3 x pi/2

or 3x^2 + x pi/2 <= 4x pi/2=2x pi.

So 2x>=(3x^2 + x pi/2)/pi ------------- equation 2.

Adding eq 1 and 2,

sin x + 2x>(3-pi/2)x/pi + (3x^2 + x pi/2)/pi = (3x^2+3x)/pi in the interval (0, pi/2]

As far as x=0 is concerned, the equality holds at x=0, with LHS and RHS both equal to 0.

So sin x + 2x >= (3x^2+3x)/pi in the interval [0, pi/2]

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