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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 real 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 $x \u003c z \u003c y$.\n3. The [[exponential function|0RMVS7RXZX]] is differentiable on $\\mathbb{R}$ with [[derivative $e^x$|0EX9GFJ0B3]]. So it is also [[continuous|07ZMX4ANJS]].\n4. By the [[mean value theorem|0BARQY52N5]] on $[x, z]$, there is $c_1 \\in (x, z)$ with $e^z - e^x = e^{c_1} (z - x) = e^{c_1} (1 - t)(y - x)$.\n5. By the mean value theorem on $[z, y]$, there is $c_2 \\in (z, y)$ with $e^y - e^z = e^{c_2} (y - z) = e^{c_2} \\, t (y - x)$.\n6. Since $c_1 \u003c z \u003c c_2$ and the [[exponential function is strictly increasing|0KN2V7840E]], we have $e^{c_1} \u003c e^{c_2}$.\n7. Since $t + (1 - t) = 1$, we can write\n   $$t e^x + (1 - t) e^y - e^z = (1 - t)(e^y - e^z) - t (e^z - e^x).$$\n8. By steps 4 and 5, the right side equals $t (1 - t)(y - x)\\big(e^{c_2} - e^{c_1}\\big)$.\n9. Each factor is positive, by step 6 and since $0 \u003c t \u003c 1$ and $x \u003c y$. So $e^{t x + (1 - t) y} \u003c t e^x + (1 - t) e^y$.\n10. This holds for all $x \\ne y$ and $t \\in (0, 1)$. So the exponential function is [[strictly convex|06BA5JZKTQ]] on $\\mathbb{R}$.\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 $e^x$ inside that piece. Since $e^x$ is strictly increasing, the right slope is larger than the left slope, so the graph bends upward and lies strictly below the chord.\n",
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