[proofplan] The point $b$ lies between $a$ and $c$, so it can be written as a convex combination of $a$ and $c$. Applying convexity of $f$ to this representation gives a single inequality relating $f(a)$, $f(b)$, and $f(c)$. Rearranging that inequality in two different ways yields the two desired comparisons of secant slopes. [/proofplan]

[step:Express $b$ as a convex combination of $a$ and $c$] Define \begin{align*} t := \frac{b-a}{c-a}. \end{align*} Since $a<b<c$, we have $c-a>0$ and $0<t<1$. Moreover, \begin{align*} (1-t)a+tc=b. \end{align*} Because $I$ is an interval and $a,c\in I$, every point between $a$ and $c$ lies in $I$; in particular, this convex combination is an element of $I$. [/step]

[step:Apply convexity at the point $b$]Since $f:I\to\mathbb{R}$ is convex, $a,c\in I$, and $t\in[0,1]$, convexity gives \begin{align*} f((1-t)a+tc)\le (1-t)f(a)+t f(c). \end{align*} Using $(1-t)a+tc=b$ and substituting the value of $t$, this becomes \begin{align*} f(b)\le \frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c). \end{align*}[/step]

[guided]The key point is that convexity controls the value of $f$ at an interior point by the corresponding linear interpolation of the endpoint values. We have already written $b$ in the form \begin{align*} b=(1-t)a+tc \end{align*} with \begin{align*} t=\frac{b-a}{c-a}. \end{align*} Because $a<b<c$, the denominator $c-a$ is positive and the numerator $b-a$ is strictly between $0$ and $c-a$, so $0<t<1$. Thus the convexity inequality for the map $f:I\to\mathbb{R}$ applies to the points $a,c\in I$ and the parameter $t\in[0,1]$. Convexity gives \begin{align*} f((1-t)a+tc)\le (1-t)f(a)+t f(c). \end{align*} Since $(1-t)a+tc=b$, the left-hand side is $f(b)$. Also, \begin{align*} 1-t=1-\frac{b-a}{c-a}=\frac{c-b}{c-a}. \end{align*} Substituting these identities into the convexity inequality gives \begin{align*} f(b)\le \frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c). \end{align*} This is the single inequality from which both secant slope comparisons will be extracted.[/guided]

[step:Rearrange the convexity inequality to compare $S_f(a,b)$ and $S_f(a,c)$] From \begin{align*} f(b)\le \frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c), \end{align*} subtract $f(a)$ from both sides: \begin{align*} f(b)-f(a)\le \frac{b-a}{c-a}(f(c)-f(a)). \end{align*} Since $b-a>0$, division by $b-a$ preserves the inequality and gives \begin{align*} \frac{f(b)-f(a)}{b-a}\le \frac{f(c)-f(a)}{c-a}. \end{align*} By the definition of the secant slope, this is \begin{align*} S_f(a,b)\le S_f(a,c). \end{align*} [/step]

[step:Rearrange the same inequality to compare $S_f(a,c)$ and $S_f(b,c)$] Again starting from \begin{align*} f(b)\le \frac{c-b}{c-a}f(a)+\frac{b-a}{c-a}f(c), \end{align*} subtract $f(c)$ from both sides: \begin{align*} f(b)-f(c)\le \frac{c-b}{c-a}(f(a)-f(c)). \end{align*} Multiplying by $-1$ reverses the inequality: \begin{align*} f(c)-f(b)\ge \frac{c-b}{c-a}(f(c)-f(a)). \end{align*} Since $c-b>0$, division by $c-b$ preserves the inequality and gives \begin{align*} \frac{f(c)-f(b)}{c-b}\ge \frac{f(c)-f(a)}{c-a}. \end{align*} Equivalently, \begin{align*} S_f(a,c)\le S_f(b,c). \end{align*} Combining this inequality with $S_f(a,b)\le S_f(a,c)$ proves \begin{align*} S_f(a,b)\le S_f(a,c)\le S_f(b,c). \end{align*} [/step]