Can someone explain the detail of the proof (rectifiable curve)

Question

Curve $C$ in the plain, \begin{align} C:= \begin{cases} x=\phi(t) \\ y=\psi(t) \end{cases} \quad, \quad a \le t \le b \end{align}

The graph of C is $\{(x, y): x=\phi(t), y=\psi(t), a\le t\le b\}$

$\Gamma=\{t_0=a, t_1, \cdots, t_m=b\}$ is a partition of [a, b]

The corresponding points $P_i=\left(\phi(t_i), \psi(t_i)\right), i=0, \cdots, m$ on the graph $C$.

$l$ is the length, $$ l(\Gamma)=\sum_{i=1}^m \sqrt{\left[\phi(t_i)-\phi(t_{i-1})\right]^2 \left[\psi(t_i)-\psi(t_{i-1}) \right]^2} $$

$L$ is the length of the curve $C$, $$L=L(C)=\sup_\Gamma l(\Gamma)$$

If $\lim_{|\Gamma|\to0}l(\Gamma)$ exists, then $C$ is called rectifiable curve.

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Theorem

$C$ is rectifiable if and only if both $\phi$ and $\psi$ are of bounded variation. Moreover, $$V(\phi), V(\psi)\le L \le V(\phi)+V(\psi)$$

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Proof

1. for real $a$, $b$,$$|a|, |b| \le \sqrt{a^2+b^2} \le |a|+|b|$$
2. If $C$ is rectifiable and $\Gamma=\{t_i\}$ is a partition of $[a, b]$, then the inequality of l $$l(\Gamma)=\sum_i \sqrt{\left[\phi(t_i)-\phi(t_{i-1})\right]^2+\left[\psi(t_i)-\psi(t_{i-1})\right]^2}\le L$$ implies $$\sum |\phi(t_i) - \phi(t_{i-1})| \le L \quad \text{and} \quad \sum |\psi(t_i) - \psi(t_{i-1})| \le L$$
3. Hence, $V(\phi), V(\psi) \le L$
4. On the other hand, for any $C$, $$l(\Gamma) \le \sum |\phi(t_i) - \phi(t_{i-1})| + \sum |\psi(t_i) - \psi(t_{i-1})| \le V(\phi) + V(\psi)$$
5. Hence, $L \le V(\phi) + V(\psi)$
6. By 3 and 6 process, $$V(\phi), V(\psi)\le L \le V(\phi)+V(\psi)$$

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I do not understand process 5. How can we get $L \le V(\phi) + V(\psi)$? which one is correct? $$l(\Gamma) \le \sum |\phi(t_i) - \phi(t_{i-1})| + \sum |\psi(t_i) - \psi(t_{i-1})| \le L \le V(\phi) + V(\psi)$$ $$l(\Gamma) \le L \le \sum |\phi(t_i) - \phi(t_{i-1})| + \sum |\psi(t_i) - \psi(t_{i-1})| \le V(\phi) + V(\psi)$$

Can you explain why it does?