I can prove the result with the restriction $f(x)\geq0$ on $[0,1]$.
WLOG we may take $a=0$, $b=1$ since applying the result to $g(x)=f((b-a)x+a)$ proves the result in the general case.
The idea is that the condition $\max\limits_{0\leq x\leq 1} \left|f'(x)\right| <K$ and $f(0)=f(1)=0$ implies that $f$ lies in the triangle $\{(0,0),(1/2,k/2),(1,0)\}$ which has area $K/4$ and hence $\int_{0}^{1}f(t)dt<K/4. $
Proof:
$f'(x)\leq\left|f'(x)\right| <K$ implies by integrating both sides from $0$ to $x<1/2$ that $f(x)<Kx$ for $x\in[0,1/2]$. This is valid as $f$ has a continuous derivative. Integrating again gives $\int_0^xf(t)dt<Kx^2/2.$ and hence $\int_0^{1/2}f(t)dt<K/8.$ Repeating for $g(x)=f(1-x)$ gives $\int_{1/2}^{1}f(t)dt<K/8.$ Add to obtain $\int_{0}^{1}f(t)dt<K/4.\blacksquare$
Setting $K=\frac{1}{4}\int_0^1f(x)dx$ we have $\int_0^1f(x)dx<K/4=\frac{1}{4}\max\limits_{0\leq x\leq 1} \left|f'(x)\right|$ and hence $$\max\limits_{0\leq x\leq 1} \left|f'(x)\right|>\frac{1}{4}\int_0^1f(x)dx.$$ This is a contradiction. Therefore $\max\limits_{0\leq x\leq 1} \left|f'(x)\right| \geq K=\frac{1}{4}\int_0^1|f(x)|dx.$ This proves the result on the assumption that $f(x)\geq0$ on $[0,1]$.
Best Answer
Let $f (t)=1+2\ln(t)-t^2$ for $t>0$.
for $x>0$ , $f $ is continuous at $[x,1] $ or $[1,x]$ and differentiable at $(x,1) $ or $(1,x) $ thus by MVT, exists $c$ strictly between $x $ and $1$ such that
$$f (x)-f (1)=(x-1)f'(c) $$ $$1+2\ln (x)-x^2=2(x-1)(1/c -c) $$ $$=2 (x-1)\frac {1-c^2}{c} $$
If $x>1$ then $1-c^2<0$ and
if $x <1$ then $1-c^2>0$
thus in all cases $$f (x)\le 0.$$