From derivative of $f$ to squaring $f$ I am dealing with a math problem which I can't resolve.
Given $f:\mathbb{R}\to\mathbb{R}$ twice differentiable with the following property:
$$ 2x f(x)\geq f'(0)-f'(x)$$
Show that
$$ f''(0)+f^2(0)\geq -1$$
I've been trying to use the derivative definition:
$$\lim \limits_{x \to x_0} \frac{f(x) -f(x_0)}{x-x_0}$$
But I got no result.
 A: $f''(0)$ exists because $f$ is twice differentiable.
By definition
$f''(0)=\lim\limits_{x\rightarrow0}\frac{f'(x)-f'(0)}{x}$
You can minorate the fraction using the given property $2xf(x)\geq f'(0)-f'(x)$ :
for $x>0$,$\frac{f'(x)-f'(0)}{x}\geq-2f(x)$
The LHS has a limit in $0$ because $f$ is twice differentiable. Taking the limits of continuous functions preserves inequalities, hence:
$f''(0)\geq-2f(0)$
From there 
$f''(0)+f^2(0)\geq-2f(0)+f^2(0)=(1-f(0))^2-1$
And you can conclude using that a square is always non-negative.
A: hint
For $x>0,$
$$-2f(x)\le \frac{f'(x)-f'(0)}{x}$$
thus
$$f''(0^+)\ge - 2f(0)$$
for $x<0$,
$$-2f(x)\ge \frac{f'(x)-f(0)}{x}$$
$$ \implies    f''(0^-)\le -2f(0)$$
so $$f''(0)=-2f(0)$$
finally
$$-2f(0)+f^2(0)+1=(f(0)-1)^2\ge 0$$
A: Note that
$$ f'(x)-f'(0)\ge-2xf(x) $$
and hence
$$ \frac{f'(x)-f'(0)}{x}\ge-2f(x), \forall x>0 $$
and
$$ \frac{f'(x)-f'(0)}{x}\le-2f(x), \forall x<0. $$
So
$$ f''_+(0)=\lim_{x\to0^+}\frac{f'(x)-f'(0)}{x}\ge-2f(0) \tag{1}$$
and
$$ f''_-(0)=\lim_{x\to0^-}\frac{f'(x)-f'(0)}{x}\le-2f(0). \tag{2}$$
From (1)(2), one has
$$ f''(0)=-2f(0) $$
and hence
$$ f''(0)+f^2(0)=-2f(0)+f^2(0)=(f(0)-1))^2-1\ge-1. $$
