Proving the inequality $2 \cos \phi + \phi \sin \phi -2 \leq 0$ How to prove the following inequality
$$
2 \cos \phi + \phi \sin \phi -2 \leq 0
$$
for $0<\phi<\pi/2$.
 A: As shown in this answer, on the given range,
$$
\phi/2\le\tan(\phi/2)
$$
multiply both sides by $2\sin(\phi)$ to get
$$
\phi\sin(\phi)\le2\tan(\phi/2)\sin(\phi)=2-2\cos(\phi)
$$
A: $2(\cos \phi-1)=2(1-2\sin^2 (\phi/2)-1)=2(-2\sin^2\dfrac{\phi}{2})$
We need to show that 
$\phi \sin \phi \le 4\sin^2\dfrac{\phi}{2}$ 
$2\phi \sin \dfrac{\phi}{2} \cos \dfrac{\phi}{2} \le 4\sin^2\dfrac{\phi}{2}$
$\phi \cos \dfrac{\phi}{2} \le 2 \sin \dfrac{\phi}{2}$
Since, $0<\phi<\dfrac{\pi}{2} \implies 0 < \dfrac{\phi}{2}< \dfrac{\pi}{4} $
Hint: What can you say about the graph of $\cos \phi $ as compared to $\sin \phi$ in the region $\{0, \dfrac{\pi}{4} \}$?
A: Apply the mean value theorem to $\tan x$ on $[0,x]$:
$$\tan x-\tan 0=(1+\tan^2 c)(x-0)$$ where $c\in ]0,x[$
Since $c>0$ then $1+\tan^2 c>1$ so $\tan x\geq x$
Now put $x=\frac{\phi}{2}$ and multiply by $\sin\phi=2\cos\frac{\phi}{2}\sin\frac{\phi}{2}$:
$$2\cos\frac{\phi}{2}\sin\frac{\phi}{2}\tan\frac{\phi}{2}\geq 2\frac{\phi}{2}\cos\frac{\phi}{2}\sin\frac{\phi}{2}$$
$$2\sin^2\frac{\phi}{2}\geq\frac{\phi}{2}\sin\phi$$
$$1-\cos\phi\geq\frac{\phi}{2}\sin\phi$$
which is what you want.
A: You can rearrange the LHS to get
$$(x \cos{x}-\sin{x}) 2\sin{x}$$
Where $x=\phi/2$.  By expressing the quantity in parentheses in a Taylor series, you can see that it is always less than 0.
