Showing that $4f(x)\le x+2, \forall x\ge 0$ with the help of $f''$ and the tangent of $C_f$

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Given $$f(x)={e^x \over e^x+1}, x\in \mathbb R$$$f$ strictly increasing at $\mathbb R$

$f''(x)>0, (-\infty,0]$ and $f''(x)<0, [0,+\infty)$

•Turning point $A(0,{1\over 2})$

I) $a)$ Find the equation of the tangent of $C_f$ at the turning point of $C_f$ and then, $b)$ show that: $$4f(x)\le x+2, \forall x\ge0$$

Personal work:

I) $a)$ The equation of the tangent at the point $A(0,{1\over 2})$ is:

$y-f(0)=f'(0)(x-0)\iff\cdots\iff y={1 \over 4}x+{1 \over 2}$

$b)$ Am I allowed to use the equal sign($=$) or should I use this one: $\iff$ ?

Example 1 $$\forall x \ge 0 : 4f(x) \le x+2 \iff _{4>0} f(x) \le {x+2 \over 4}={x \over 4}+{2 \over 4}={x \over 4} + {1 \over 2} = y \iff f(x)-y \le 0 \iff {e^x \over e^x+1}-{x \over 4}+{1 \over 2}\le0 \dots=?$$

Example 2 $$\forall x \ge 0 : 4f(x) \le x+2 \iff _{4>0} f(x) \le {x+2 \over 4}\iff f(x)\le{x \over 4}+{2 \over 4} \iff f(x) \le {x \over 4} + {1 \over 2} \iff f(x)\le y \iff f(x)-y \le 0 \iff {e^x \over e^x+1}-{x \over 4}+{1 \over 2}\le0 \dots=?$$

Which one of the two examples is mathematically correct as for the signs ($=, \iff$) ?