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  "body": "\n## Proof\n\nWe apply the mean value theorem on two adjacent intervals.\n\n1. Let $x \\ne y$ be in $(0, \\infty)$ and let $t \\in (0, 1)$. If we swap $x$ with $y$ and $t$ with $1 - t$, both sides of the claimed inequality stay the same. So we may assume $x \u003c y$.\n2. Let $z = t x + (1 - t) y$. Then $z - x = (1 - t)(y - x) \u003e 0$ and $y - z = t (y - x) \u003e 0$. So $0 \u003c x \u003c z \u003c y$.\n3. The [[natural logarithm|0044CPHJBV]] is differentiable on $(0, \\infty)$ with [[derivative $1/x$|07X1QA6BRW]]. So it is also [[continuous|07ZMX4ANJS]] there.\n4. By the [[mean value theorem|0BARQY52N5]] on $[x, z]$, there is $c_1 \\in (x, z)$ with $\\ln z - \\ln x = \\frac{z - x}{c_1} = \\frac{(1 - t)(y - x)}{c_1}$.\n5. By the mean value theorem on $[z, y]$, there is $c_2 \\in (z, y)$ with $\\ln y - \\ln z = \\frac{y - z}{c_2} = \\frac{t (y - x)}{c_2}$.\n6. Since $0 \u003c c_1 \u003c z \u003c c_2$, we have $\\frac{1}{c_1} \u003e \\frac{1}{c_2}$.\n7. Since $t + (1 - t) = 1$, we can write\n   $$\\ln z - t \\ln x - (1 - t) \\ln y = t (\\ln z - \\ln x) - (1 - t)(\\ln y - \\ln z).$$\n8. By steps 4 and 5, the right side equals $t (1 - t)(y - x)\\big(\\frac{1}{c_1} - \\frac{1}{c_2}\\big)$.\n9. Each factor is positive, by step 6 and since $0 \u003c t \u003c 1$ and $x \u003c y$. So $\\ln(t x + (1 - t) y) \u003e t \\ln x + (1 - t) \\ln y$.\n10. This holds for all $x \\ne y$ in $(0, \\infty)$ and all $t \\in (0, 1)$. So $\\ln$ is strictly [[concave|0EF2VG8XKQ]] on $(0, \\infty)$.\n\n## Idea\n\nThe point $z$ splits $[x, y]$ into two pieces. By the mean value theorem, the slope of the graph over each piece is a value of $1/x$ inside that piece. Since $1/x$ is strictly decreasing, the right slope is smaller than the left slope, so the graph bends downward and lies strictly above the chord.\n",
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