Proving $\arctan x > \frac x{1+x^2}, \forall x >0$ with a helper function 
Prove $\arctan x > \frac x{1+x^2}, \forall x >0$

There's the approach using Lagrange's, but is it also possible to define a function like so? 
$f(x)=\arctan x - \frac x{1+x^2}$, take the derivative:
$f'(x)=\frac {2 x^2} {(1+x^2)^2}$, if $f'(x)=0$ then $x=0$.
From the second derivative test we get that $f''(0)=0$ so $x=0$ isn't an extrema and from Rolle's we know that there's at most only one solution for $f(x)$, we know that the only solution is at $x=0$ so we can conclude that $f(x)>0, \forall x>0$.
Is this a valid way?
 A: Your proof is not valid because $f''(0) = 0$ does not imply that $x = 0$ 
is not an extremum. As already said in the other answers, one can easily see
that $f' > 0$ and therefore $f$ is strictly increasing.
But you can prove the inequality also by applying the Mean Value Theorem
directly to $\arctan$, without a helper function. For $x > 0$:
$$
 \arctan(x) = \arctan(0) + (x- 0)\arctan'(c)  \quad \text {for some } c \in (0, x) \\
  =  \frac {x}{1 + c^2} > \frac {x}{1 + x^2} \, .
$$
A: I think the easiest way to prove the inequality is to consider that:
$$\arctan x=\int_{0}^{x}\frac{dt}{1+t^2}>\int_{0}^{x}\frac{dt}{1+x^2}=\frac{x}{1+x^2}.$$
A: This way is not valid. If I unterstood you correctly, your argumentation would work with $-f$ as well. Just show $$ f(0) = 0,\quad f'(x) > 0 \quad \forall x > 0$$ and then you can use the Mean Value Theorem to show that for $y > 0$
$$ (f(y) - f(0))\cdot y = f'(\xi)\quad \xi \in [0,y]$$ 
and therefore $f(y) > 0$.
A: You have a good start! Observe this $f(0)=0$ and $f'(x) > 0,\, x > 0 $ which means the function is strictly increasing then $f(x)\geq 0,\, x\geq 0$ and the inequality follows. 
