Is there a geometrical method to prove $x<\frac{\sin x +\tan x}{2}$? Suppose $x \in (0,\frac{\pi}{2})$
and we want to prove $$x<\frac{\sin x +\tan x}{2}$$I tried to prove it by taking $f(x)=\sin x+ \tan x -2x$ and show $f(x) >0 ,when\space  x \in (0,\frac{\pi}{2})$ take f'$$f'=\cos x +1+\tan ^2 x-2\\=\tan^2 x-(1-\cos x)\\=\tan ^2 x-2sin^2(\frac x2)$$ I get stuck here ,because the last line need to be proved $\tan ^2 x>2sin^2(\frac x2) ,when\space  x \in (0,\frac{\pi}{2})$
$\bf{Question}:$ Is there a geometrical method to prove the first inequality ? (or other idea)
Thanks in advance.  
$\bf{Remark}: $I can see the function is increasing $when\space  x \in (0,\frac{\pi}{2})$ like below :https://www.desmos.com/calculator/www2psnhmu
 A: let $$f(x)=\sin(x)+\tan(x)-2x$$ then $$f(0)=0$$ and $$f'(x)=\cos(x)+\frac{1}{\cos(x)^2}-2$$ and this can be written as $$f'(x)=\frac{(1-\cos(x))(1-\cos(x)^2)+\cos(x)(1-\cos(x))}{\cos(x)^2}>0$$
A: 
We have $\tan x\gt x$  for $x\in(0,\frac\pi2)$ the result follows,  Using AM-GM-HM inequalites we have , 
  $$\color{blue}{\frac{\sin x+ \tan x}{2} \ge \sqrt{\sin x\tan x} \ge \frac{2}{\frac{1}{\sin x }+\frac{1}{\tan x} } = 2\tan \frac x2 \gt x}$$

Indeed, $$ \frac{1}{\sin 2u }+\frac{1}{\tan 2u}= \frac{1}{\sin 2u }+ \frac{\cos 2u }{\sin 2u } =\frac{2\cos^2 u}{2\cos\sin u} = \frac{1}{\tan u}$$
A: For any $x$ in the given interval
$$\sin(x)+\tan(x)=\int_{0}^{x}\left(\cos(u)+\frac{1}{\cos^2(u)}\right)\,du\stackrel{AM-GM}{\geq}2\int_{0}^{x}\frac{du}{\sqrt{\cos(u)}} $$
and the last integral is clearly $>2\int_{0}^{x}1\,dx = 2x$.
The same approach proves the stronger, non-trivial inequality

$$ \forall x\in\left(0,\tfrac{\pi}{2}\right),\qquad \color{red}{2}\sin(x)+\tan(x)>\color{red}{3x}.$$


Here it is a properly geometric proof.

Given a circle sector with amplitude $2\theta$, we may consider the associated arc and the parabola through the midpoint and the endpoints of such arc. The union of the parabolic segment and the triangle with side lengths $1,1,2\sin\theta$ is a region strictly contained in the circle sector. The area of the parabolic segment is $\frac{4}{3}$ of the area of the shaded triangle, hence we have:
$$\cos\theta\sin\theta+\frac{4}{3}\sin\theta(1-\cos\theta)< \theta.$$
A: Here's a geometric argument, but it isn't as slick as some of the Calculus-based ones.

Consider the unit circle about $O$, through $R$ and $S$, with $\theta = \angle ROS$. The perpendicular from $S$ to $\overline{OR}$ has length $\sin\theta$, while the perpendicular from $R$ up to $T$ on the extension of $\overline{OS}$ has length $\tan\theta$. Let $M$ be the midpoint of $\overline{ST}$.
Then 
$$2\;|\text{area of sector}\;ROS| = \theta \qquad\text{and}\qquad 2\;|\triangle ORM| = \frac{1}{2}\left(\sin\theta + \tan\theta\right)$$
"All we need to do" is show that the triangle has more area than the sector. This seems pretty clear; after all, the triangle contains almost-all of the sector, except for the circular segment defined by $\overline{KR}$, where $K$ is the intersection of $\overline{RM}$ and the circle. There is a concern, though, that the excess area in the triangular region $KSM$ could be less than that of the tiny sliver of a circular segment for small $\theta$; we need to dispel that concern.
There's probably a simpler route to this, but I coordinatized and, with the help of Mathematica, found
$$M = \left(\frac{1 + \cos\theta}{2}, \frac{\sin\theta (1 + \cos\theta)}{2 \cos\theta}\right)$$
$$K = \left(\frac{1 + 3 \cos\theta + 2 \cos^2\theta + 2 \cos^3\theta}{1 + 3 \cos\theta + 4 \cos^2\theta}, \frac{2 \sin\theta \cos\theta ( 1 + \cos\theta)}{1 + 3 \cos\theta + 4 \cos^2\theta}\right)$$
so that (after a bit more symbol-crunching)
$$\frac{|\overline{MK}|}{|\overline{KR}|} = \frac{1 + 3 \cos\theta}{4 \cos^2\theta} = 1 + \frac{1 + 3 \cos\theta - 4 \cos^2\theta}{4 \cos^2\theta} = 1 + \frac{(1-\cos\theta)(1 + 4 \cos\theta)}{4 \cos^2\theta} > 1$$
for $0 < \theta < \pi/2$.
This says that $\overline{MK}$ is longer than $\overline{KR}$, so that we could reflect $R$ in $K$ to get $R^\prime$, and copy circular segment $KR$ as circular segment $KR^\prime$ inside $\triangle ORM$ yet tangent to the unit circle (and therefore outside of sector $ORS$). 

Consequently, the triangle definitely has more area than the sector, so we're done. $\square$
A: Let $AB=AC=1$ and $\angle BAC = \alpha$. Let $D \in AC$ with $DB \perp AB$. Let $E \in AB$ with $CE \perp AB$. Let $F$ and $G$ be the midpoints of $EB$ and $CD$, respectively. Let $H \in BD$ with $CH \perp AD$. Let $I \in FG$ with $IH \perp GF$. 

Clearly $IH < GD = CG$, therefore 
$$IG^2 = GH^2 - IH^2 > GH^2 - CG^2 = CH^2.$$
Therefore $GI>CH$. It follows that 
\begin{align*}
\frac{\sin \alpha + \tan \alpha}{2} & = \frac{CE+BD}{2} = GF = GI+IF 
> CH+IF = CH+BH = 2BH \\
&= 2\tan \frac \alpha2 > 2 \cdot \frac \alpha2 = \alpha.\end{align*}
A: Here's a far cleaner argument than in my previous answer.

Considering acute (and non-zero) $\theta$ ... 

In the figure, 
$\overline{PS}$ is a leg of a right triangle with hypotenuse $\overline{PT}$. Thus,
$$|PS| < |PT| \quad\implies\quad |PR| < |PT| \quad\implies\quad |PR| < \frac12|TR| \tag{$\star$}$$
(FYI, this proves the convexity of the tangent function: $\tan\frac12\theta < \frac12\tan\theta$.) Continuing, with "cseg" indicating "circular segment", ...
$$\begin{align}
|\text{cseg}RS| \;<\; |\triangle PRS| &\;\stackrel{\star}{<}\; \frac12|\triangle TRS| = \frac12\left(\;|\triangle TRO|-|\triangle SRO|\;\right) \\[6pt]
\implies\quad|\text{sect}RS| \;=\; |\text{cseg}RS| + |\triangle SRO| &\;<\; \frac12\left(\;|\triangle TRO|+|\triangle SRO|\;\right) \\[6pt]
\implies\quad \frac12\cdot 1^2\cdot\theta &\;<\; \frac12\left(\;\frac12\cdot 1\cdot \tan\theta \;+\; \frac12\cdot 1\cdot\sin\theta\;\right)
\end{align}$$
and the result follows. $\square$
A: $$f=\sin (x)+\tan (x)-2 x\implies f'=\cos (x)+\sec ^2(x)-2$$ Now, using the tangent half-angle substitution, we have $$f'=\frac{t^2}{4}+\frac{2}{\sqrt{t^2+4}}-1$$ Squaring leads to $$f'=0 \qquad \text{if} \qquad -\frac{t^6}{16}+\frac{t^4}{4}+t^2=0$$ the real roots are $$t=\pm\sqrt{2 \left(1+\sqrt{5}\right)}$$ Considering the positive root then $$f'=0 \qquad \text{if} \qquad x=2\tan ^{-1}\left(\sqrt{2 \left(1+\sqrt{5}\right)}\right)\approx 2.39255$$ So, the derivative does not cancel in the interval and it is always non negative.
Funny would also be a Taylor expansion; up to any order, the coefficients are positive.
A: From $f' = \tan^2 x-(1-\cos x)$, we get $f'' = 2\tan x \sec^2 x - \sin x = \sin x(2\sec^3 x -1) > 0$ in $\left(0, \frac{\pi}{2}\right)$. Thus $f'$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and thus $f'(x) > f'(0) = 0$. Thus $f$ is increasing.
A: We've $\dfrac{(\sin x + \tan x)}2\ge \sqrt{\sin x\tan x}$
(AM GM Inequality)
Equality holds when $x=0$
Using expansions of $\sec x$ and $\cos x$ one can easily show that $\sin x \tan x>x^2$
Hence the result
A: In fact one can prove a better inequality 
$$ \frac { 2 \sin x + \tan x }{3} > x$$ for $x \in (0, \frac{\pi}{2})$.
Indeed, the difference $\frac { 2 \sin x + \tan x }{3} - x$ has derivative
$$\frac{1}{3 \cos^2 x} ( 2 \cos^3 x - 3 \cos ^2 x + 1 )$$ and since the expression $2 t^3 - 3 t^2 + 1 = (t-1)^2 ( 2 t+ 1)$ is $>0$ on $[0,1)$, the function in $x$ is strictly increasing on $[0, \pi/2]$.
Obs: One can check this is the "best" inequality of its kind.
$\bf{Added}:$ Like @Claude Leibovici: noticed for the original question,  in this case also the Maclaurin expansion of the difference $2 \sin x + \tan x - 3 x$ has all the coefficients positive. This can be checked by using the explicit Taylor- Maclaurin expansions of $\sin$ and $\tan$. 
