How to prove 4cos⁡(x)+xsin⁡(x)≥4−x2 for all real x, with equality if and only if x=0?

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Juan Goodwin · Accepted Answer

Step 1Let&amp;nbsp;h(x)=4cos⁡(x)+xsin⁡(x)+x2−4; note that h(x) is even. For reference, we haveh&amp;nbsp;′(x)=2x+xcos⁡(x)−3sin⁡(x)h&amp;nbsp;″(x)=2−2cos⁡(x)−xsin⁡(x)h(3)(x)=sin⁡(x)−xcos⁡(x)h(4)(x)=xsin⁡(x)h(5)(x)=sin⁡(x)+xcos⁡(x)h(6)(x)=2cos⁡(x)−xsin⁡(x)Claim: It suffices to show&amp;nbsp;h(x)≥0&amp;nbsp;on&amp;nbsp;[0,&amp;nbsp;π]We have&amp;nbsp;h′(x)=x(1+cos⁡(x))+(x−3sin⁡(x))For&amp;nbsp;x&amp;gt;0&amp;nbsp;the first term is non-negative and for&amp;nbsp;x&amp;gt;π&amp;nbsp;the second one iss as well the unique positive real solution of&amp;nbsp;x=3sin⁡(x)&amp;nbsp;is&amp;nbsp;x≈2.28&amp;nbsp;but we take&amp;nbsp;x=π&amp;nbsp;for convenience). In other words, for&amp;nbsp;x&amp;gt;π&amp;nbsp;we have&amp;nbsp;h′(x)&amp;gt;0&amp;nbsp;and&amp;nbsp;h(π)&amp;gt;0.Claim:&amp;nbsp;h′(0)=0&amp;nbsp;(obvious) and h has its global minimum at zero.This is basically a corollary to the Second Derivative Test. Note that&amp;nbsp;h&#039;&#039;&#039;&#039;(x)≥0&amp;nbsp;on&amp;nbsp;[−π,π]&amp;nbsp;as the signs of the terms agree, which implies&amp;nbsp;h&amp;nbsp;″&amp;nbsp;has a minimum at&amp;nbsp;x=0, i.e. that&amp;nbsp;h&amp;nbsp;″(x)≥0&amp;nbsp;on&amp;nbsp;[−π,π]. Alternatively, since the sixth derivative is the first to evaluate to a nonzero number at zero and this value is positive, h has a minimum at&amp;nbsp;x=0.

unduncjineei5r3 · Accepted Answer

Step 1By Taylor&#039;s theorem with the mean-value forms of the remainder,1) f(x)=x2+4cos⁡x+xsin⁡x−4=x6360+(−3sin⁡(ζ)−ζcos⁡(ζ))x75040where ζ is between 0 and x, andf(7)(x)=−3sin⁡(x)−xcos⁡xSo if |x| is small enough (e.g., |x|&amp;lt;11) (1) is non-negative. If |x|&amp;gt;11 (11 can be made smaller), we can easily prove x2+4cos⁡x+xsin⁡x−4&amp;gt;0.

How to prove \(\displaystyle{4}{\cos{{\left({x}\right)}}}+{x}{\sin{{\left({x}\right)}}}\geq{4}-{x}^{{{2}}}\) for all real

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