Prove that, $f'(0) \ge -\sqrt{2}$ for a function $f$ satisfying some conditions on $(-1,1)$. 
Let $f:(-1,1)\to \mathbb{R}$ be a twice differentiable function such that, $f(0)=1$, $f'(x)≤0$, $f(x)≥0$ and $f''(x)≤f(x)$ for all $x≥0$. Prove that, $f'(0)≥-\sqrt2$

Progress: I was able to prove $f'(0)≥-2$. For this I have applied Cauchy MVT for $f'$ and $g$ on $[0,x]$ for any $x\in (0,1)$ where $g: [0,x]\to\mathbb{R}$ is defined by, $g(x)=\sqrt{x+1}$. But, couldn't find another approach to get $-\sqrt{2}$. 
 A: As @JackD'Aurizio has already mentioned in comment to the other answer, a possible lower bound is:

$$f'(0) \ge -\coth 1$$

One way to get it, starting from 
$$\begin{align}f''(x) \le f(x) &\implies e^{x}(f'(x)+f''(x)) \le e^{x}(f(x)+f'(x))\\& \implies \frac{d}{dx}(e^x f'(x)) \le \frac{d}{dx}(e^xf(x))\,\,\,\textrm{ [integrating both sides from $0$ to $x$] } \\&\implies f(0) - f'(0) \le e^{x}(f(x)-f'(x)) \\&\implies e^{-2x}(1-f'(0)) \le \frac{d}{dx}(-e^{-x}f(x)) \,\,\,\textrm{ [integrating both sides from $0$ to $x$] }\\&\implies \frac{1}{2}(1-e^{-2x})(1-f'(0)) \le f(0) - e^{-x}f(x)\\& \implies 0 \le f(x) \le e^{x} + (f'(0) - 1)\sinh x\\&\implies -\coth x \le f'(0)\end{align}$$
Supremum of LHS is attained at $x = 1$ for $x \in (0,1)$.
Note that if we were to use $f'(x) \le 0$, that would further give us:
$$\sup_{x \in [0,1)} f(x) = f(0) = 1 \le \sup_{x \in [0,1)} e^{x} + (f'(0) - 1)\sinh x$$
Then I guess we'd have to check if it's an improvement on the previous bound or not.
A: Initially I had posted a solution but it was incorrect. Here's a final solution. Beautiful problem, by the way.
For any $x\in(0,1)$, we can write, for some $c\in(0,x)$ that $f(x)=f(0)+f'(0)x+\dfrac{f''(c)x^2}{2}\leq 1+f'(0)x+\dfrac{f(c)x^2}{2}\leq1 + f'(0)x +\dfrac{x^2}{2}$ since $f(x)\leq1$ for $x\in(0,1)$.
Now using $f''(x)\leq f(x)$ we see that $f''(x)\leq f'(0)x + \dfrac{x^2}{2}+1$. Integrating both sides from $0$ to $x$ we have,
$f'(x)-f'(0)\leq \dfrac{f'(0)x^2}{2}+x+\dfrac{x^3}{6}$ and again integrate from $0$ to $x$ for any $x\in(0,1)$. We get $f'(0)(x+\dfrac{x^3}{6})\geq-\dfrac{x^2}{2}-\dfrac{x^4}{24}+f(x)-f(0)\geq-\dfrac{x^2}{2}-\dfrac{x^4}{24}-1$
Finally we have $$f'(0)\geq \dfrac{-\dfrac{x^2}{2}-\dfrac{x^4}{24}-1}{x+\dfrac{x^3}{6}}$$for any $x\in(0,1)$.
Note that the function on the right side is increasing and achieves its maximum at $x=1$, where its value is $\dfrac{-37}{28}>-\sqrt{2}$.
Hence $f'(0)\geq \sup_{x\in(0,1)}\dfrac{-\dfrac{x^2}{2}-\dfrac{x^4}{24}-1}{x+\dfrac{x^3}{6}}=\dfrac{-37}{28}>-\sqrt{2}$.
