Show inequality $\sin x + \ln(1+x)+e^x>3x$, when $x>0$ How to show, that $e^{x} - \sin x -  \frac{1}{(1+x)^{2}}  >0$ when $x>0$?
I have to show (main task), that $\sin x + \ln(1+x)+e^x>3x$, when $x>0$
I've made a function $G(x) = \sin x + \ln(1+x)+e^x-3x$
I've found derivative : $\cos x +  \frac{1}{1+x}+e^x-3$
and second derivative : $-\sin x- \frac{1}{(1+x)^2}+e^x$
and I want to show that second derivative $>0$, when $x>0$ -> so $G'(x)$ - increases,
$0<2=G'(0)<G'(x)  \rightarrow  G(x)$ increases, so  $\sin x + \ln(1+x)+e^x-3x>0 \rightarrow \sin x + \ln(1+x)+e^x>3x$.
Can I show this in that way? If yes, then please give a hint how to show
$e^{x} - \sin x -  \frac{1}{(1+x)^{2}}  >0 $
 A: By graphing the function, you can advantageously separate it in two parts.
Basically $\ln(1+x)+\sin(x)$ is positive and $e^x-3x$ has a small minimum which is well compensated by the monotonicity of the first one on the interval $\approx[0,2]$.
Let's go into the details...

*

*First we study $f(x)=\ln(1+x)+\sin(x)$
Note that $f(x)>0$ for $\ln(1+x)>1\iff x>e-1\approx 1.718$
Also $f'(x)=\dfrac 1{1+x}+\cos(x)>0$ on $[0,\frac\pi 2)$ and it will be negative somewhere in $[\frac\pi 2,\pi]$.
So $f$ is $\nearrow$ on $[0,\dfrac\pi 2)$ and since $f(0)=0$ then $f(x)\ge 0$ in this interval.
Then at some point it will start decreasing, but since $f(2)>0$ it is still positive in the interval $[\frac\pi 2,2]$, and since $e-1<2$ we conclude that $f(x)>0$ on $(0,+\infty)$.

*

*And now $g(x)=e^x-3x$
$g'(x)=e^x-3$ sign is determined in regard to $\ln(3)$, so $g$ is $\{\searrow\text{min}\nearrow\}$
Since $g(\ln(3))\approx -0.29 <0$ and limits at $\pm\infty$ is $+\infty$ then by the IVT the function has two roots.
Since $g(\frac 12)>0$ the first root $r_1$ is located in $[\frac 12,\ln(3)]$ but $f(\frac 12)\approx 0.88$ is largely sufficient to compensate the minimum of $g$ and since $f\ \nearrow$ after that then $f+g$ is positive so positive on $[r_1,\ln(3)]$
Then the second root $r_2$ is less than $2$ (which is the case since $g(2)>0$) also $f(2)>f(\frac12)$ so $f+g$ is still positive on $[\ln(3),r_2]$.
Remains the case of zero since $f(0)=0$ but $g(0)=1>0$ so $f(0)+g(0)>0$ and finally $f+g>0$ on $[0,+\infty)$.
Note: the method is not technically complicated but it requires to evaluate $f,g,f',g'$ in many points, so it is not entirely manual but requires a calculator.
