Proving $\frac{1-(e^{-2})^x}{1-e^{-2}} \ge x$, for $0 \le x \le 1$.

How to prove that, for any real number $0 \le x \le 1$, this inequality holds ? $$\frac{1-(e^{-2})^x}{1-e^{-2}} \ge x$$ I tried using wolfram alpha to solve for getting some idea, but the exact solution is very complex : here is the link.

Thank you very much.

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Prove that the function $f(x) = \dfrac{1-e^{-2x}}x$ is decreasing. Hence, $f(x) \geq f(1)$
The function $$f(x)=\frac{1-(e^{-2})^x}{1-e^{-2}}-x$$ is concave and $f(0)=f(1)=0$. Then $f(x)\geq 0$ for all $x\in[0,1]$.