$f(x)=2x-e^x<0, \forall x \in \mathbb{R}$ The question is quite simple, but I'm finding some trouble doing it...  
Prove that the function $f(x)=2x-e^x$ is negative, i.e., $f(x)<0, \forall x \in \mathbb{R}$. 
Thanks for the help.
 A: Let $f(x)=2x-\exp(x)$ then we get $$f'(x)=2-\exp(x),$$$$f''(x)=-\exp(x)<0$$
The only solution of $f'(x)=0$ is $x=\ln(2)$ and $$f(\ln(2))=2(\ln(2)-1)<0$$ thus our inequality is proven.
A: It is a convexity inequality. Since $f(x)=e^x$ is convex, its graphics lies above the graphics of the tangent line through $(\log 2,2)$, i.e.:
$$ e^x \geq 2(x-\log 2)+2 = 2x + 2(1-\log 2) > 2x.$$
A: The derivative of the above is 
$$2-e^x$$
The derivative is 0 when $x=\ln(2)$
This means the maximum is $2 \cdot (\ln(2)-1)$
However sense $x \lt 0$ the maximum is now at $x=0$, which is $-1$
This is a sketch you can fill this in with monotone properties of the functions involved. Or you could use Lagrange multipliers. 
A: The function $e^x$ is strictly increasing and $y = e\cdot x$ is a tangent to $y = e^x$ at $x = 1$, so $e^x \geq e\cdot x > 2x$ for positive $x$. For nonpositive $x$: $e^x > 0 \geq 2x$. Finally, $e^x > 2x$
A: It's clearly true for negative numbers. For non-negative numbers, considering the Taylor expansion of $e^x$, we get $2x-e^x=2x-1-x-\frac{x^2}{2}-\frac{x^3}{3!}-\dots=-1+x-\frac{x^2}{2}-\dots=-(\frac{x}{2}-1)^2-\frac{x^2}{4}-\frac{x^3}{3!}-\dots<0.$
