$0\leq f,g\in C^{1},\int_{a}^{b}\sqrt{f}\geq\int_{a}^{b}\sqrt{g}\implies\int_a^b f\geq\int_{a}^{b}g$? $f,g$ are differentiable non-negative functions on $[a,b]$ with $ \int_{a}^{b}\sqrt{f(t)}dt\geq\int_{a}^{b}\sqrt{g(t)}dt $.
So do we have that $\int_{a}^{b}f(t)dt\geq\int_{a}^{b}g(t)dt$ ? 
Does anyone have any clue about this? I tried to make a counterexample but I couldn't find. If it is true, any hint on how to prove would be appreciated.
P.S.: I hope it is true because it would make easier to find the shortest path between two points on a surface using Euler-Lagrange equation since we would have simpler differential equations to solve. If it is not true, does anyone has any tip on evaluating this paths?

No, that implication does not hold in general.
A counter-example is
$f(x) = 1$, $g(x) = 4x^2$ on the interval $[0, 1]$:
$$ 
\int_{0}^{1}\sqrt{f(t)} \, dt = 1 = \int_{0}^{1}\sqrt{g(t)} \, dt \, ,
$$
but
$$
\int_{0}^{1}f(t) \, dt = 1 < \frac 43 =  \int_{0}^{1}g(t) \, dt \, .
$$
