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  "body": "\n## Proof\n\nWe write the middle point as a convex combination of the outer points and rearrange the convexity inequality.\n\n1. Let $x \u003c y \u003c z$ be points of $I$. Put $t = \\dfrac{z - y}{z - x}$. Then $0 \u003c t \u003c 1$ and $1 - t = \\dfrac{y - x}{z - x}$.\n2. We have $t x + (1 - t) z = \\dfrac{(z - y) x + (y - x) z}{z - x} = \\dfrac{y (z - x)}{z - x} = y$.\n3. By the definition of a [[convex function|0VV0HSW1ZS]], $f(y) \\le t f(x) + (1 - t) f(z)$.\n4. Multiplying by $z - x \u003e 0$ gives $(z - x) f(y) \\le (z - y) f(x) + (y - x) f(z)$. Call this inequality $(\\ast)$.\n5. Subtract $(z - x) f(x)$ from both sides of $(\\ast)$. Since $(z - y) - (z - x) = -(y - x)$, we get $(z - x)(f(y) - f(x)) \\le (y - x)(f(z) - f(x))$.\n6. Dividing by $(z - x)(y - x) \u003e 0$ gives $\\dfrac{f(y) - f(x)}{y - x} \\le \\dfrac{f(z) - f(x)}{z - x}$.\n7. Subtract $(z - x) f(z)$ from both sides of $(\\ast)$. Since $(y - x) - (z - x) = -(z - y)$, we get $(z - x)(f(y) - f(z)) \\le (z - y)(f(x) - f(z))$.\n8. Multiplying by $-1$ reverses the inequality: $(z - x)(f(z) - f(y)) \\ge (z - y)(f(z) - f(x))$.\n9. Dividing by $(z - x)(z - y) \u003e 0$ gives $\\dfrac{f(z) - f(x)}{z - x} \\le \\dfrac{f(z) - f(y)}{z - y}$.\n10. Steps 6 and 9 together are the two claimed inequalities.\n\n## Idea\n\nThe point $y$ splits $[x, z]$ in the ratio of the weights $t$ and $1 - t$. Convexity says the graph at $y$ lies below the chord from $x$ to $z$. Seen from $x$, this makes the chord to $y$ less steep. Seen from $z$, it makes the chord from $y$ steeper.\n",
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