How to prove $4 \cos(x) + x \sin(x) \geq 4 - x^2$ for all real $x$, with equality if and only if $x=0$? 
Show $4 \cos(x) + x \sin(x) \geq 4 - x^2$ for all real $x$, with equality if and only if $x=0$.

LHS-RHS has a power series $x^6/(2×6!)−x^8/(4×8!)+x^{10}/(6×10!)−\ldots$, and I could try to prove that this is positive. But it seems messy. I was hoping for some cleverer trick, but I'm out of ideas right now.
 A: Let $h(x)=4\cos(x)+x\sin(x)+x^2-4$; note that $h(x)$ is even. For reference, we have
\begin{align}
h^\prime(x) &=2x+x\cos(x)-3\sin(x)\\
h^{\prime\prime}(x)& = 2-2\cos(x)-x\sin(x)\\
h^{(3)}(x)& =\sin(x)-x\cos(x)\\
h^{(4)}(x) &= x \sin(x)\\
h^{(5)}(x)& = \sin (x)+x \cos (x)\\
h^{(6)}(x)& =2 \cos (x)-x \sin (x)
\end{align}
Claim: It suffices to show $h(x)\ge 0$ on $[0,\pi]$.
We have $h'(x) = x(1+\cos(x)) + (x-3\sin(x))$. For $x>0$ the first term is non-negative and for $x>\pi$ the second one is as well (the unique positive real solution of $x=3\sin(x)$ is $x\approx 2.28$ but we take $x=\pi$ for convenience). In other words, for $x>\pi$ we have $h'(x)>0$ and $h(\pi)>0$.
Claim: $h'(0)=0$ (obvious) and $h$ has its global minimum at zero.
This is basically a corollary to the Second Derivative Test. Note that $h''''(x)\ge 0$ on $[-\pi,\pi]$ as the signs of the terms agree, which implies $h''$ has a minimum at $x=0$, i.e. that $h''(x)\ge 0$ on $[-\pi,\pi]$. 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 $x=0$.
A: Hint:
By Taylor's theorem with the mean-value forms of the remainder,
$$f(x)=x^2+4\cos x + x \sin x -4 = \frac{x^6}{360}+(-3\sin(\zeta)-\zeta \cos(\zeta))\frac{x^7}{5040}\tag1$$
where $\zeta$ is between $0$ and $x$, and $f^{(7)}(x)=-3 \sin (x) - x \cos x$.
So if $|x|$ is small enough (e.g., |x| < 11) $(1)$ is non-negative. If $|x|>11$ ($11$ can be made smaller), we can easily prove $x^2+4\cos x + x \sin x -4>0$. Can you take it from here?
