Prove $\frac{2 \sin x}{3}+\frac{\tan x}{3} > x$ for $x \in (0, \frac{\pi}{2})$ Please, help
Prove $\frac{2 \sin x}{3}+\frac{\tan x}{3}  > x$
$x \in (0, \frac{\pi}{2})$
 A: Letting $f(x)=2\sin x+\tan x-3x$, then differentiate it. You'll get the answer.
$$f^{\prime}(x)=2\cos x+\frac{1}{\cos^2 x}-3=\frac{2\cos^3 x-3\cos^2 x+1}{\cos^2 x}=\frac{(\cos x-1)^2(2\cos x+1)}{\cos^2 x}$$
A: Let $$f(x)=\frac23\sin x+\frac13\tan(x)-x.$$ We have $f(0)=0$. If we show that $f'(x)>0$ for $x\in(0,\pi/2)$, it follows that $f$ is increasing in this interval, and since it takes the value $0$ at $0$, it will be positive as needed.
We have $$ f'(x)=\frac23\cos x+\frac13\sec^2x-1. $$ Again, $f'(0)=0$, so it is enough to show $f''(x)>0$ for $x>0$. This shows $f'$ is increasing, so it is positive.
Finally, we have $$f''(x)=-\frac23\sin x+\frac23\sec^2x\tan x. $$ Again, $f''(0)=0$. Also, if $x\in(0,\pi/2)$, then $$f''(x)=\frac23\sin(x)(\sec^3(x)-1)>0,$$ since $\cos(x)>0$ on this interval. 
A: Let
$$f(x)=\frac23\sin x+\frac13\tan x\quad;\quad f(0)=0$$
then
$$f'(x)=\frac 2 3\cos x+\frac 1 3(1+\tan^2x)\quad;\quad f'(0)=1$$
and 
$$f''(x)=-\frac 2 3\sin x+\frac 2 3\tan x(1+\tan^2 x)\ge 0,\; x\in\left(0,\frac \pi 2\right)$$
hence $f$ is a convex function on this interval and then the curve of $f$ is above the tangent line at $x=0$ which has the equation 
$$y=f'(0)x+f(0)=x$$
Now we conclude the desired result
A: This is the optimal strengthening, $\lambda = 2$, of $\tan x \geq x$ on $(0,\pi/2)$ to $$\tan x \geq x + \lambda (x - \sin x) .$$  A moment with computer software will convince you that  $\frac{\tan x - x}{x - \sin x}$ has minimum value of $2$ at $x=0$, with positive power series coefficients of even degree, and knowing that one could write out a non-computer proof by differentiation if desired, that the inequality is both true and optimal.
A: You can differentiate both sides and show that the derivative is greater than $1$ on the given interval. Also, you know that both sides eqaul $0$ in $x=0$, and when the left hand side grows faster, it will be larger.
