- By hypothesis, $x,y$ are positive numbers, and $p,q\in(1,\infty)$ with $\frac 1p+\frac 1q=1$.
- The second derivative of the natural logarithm is negative $\log^{\prime\prime}(x)=-\frac 1{x^2} < 0$ for all positive numbers $x > 0.$
- Thus, the natural logarithm fulfills the test for concaveness.
- Since by definition $\frac 1p,\frac 1q\in(0,1),$ $\frac 1p+\frac 1q=1,$ we get by definition of concave the following inequality $$\log\left(\frac 1px+\frac 1qy\right)\ge \frac 1p\log(x)+\frac 1q\log(y).$$
- Taking the exponential function on both sides of the inequation
^{1}yields

$$\frac xp+\frac yq\ge x^{1/p}\cdot y^{1/q}.$$

^{1} Note that the exponential function is the inverse function to the logarithm.

q.e.d

| | | | created: 2020-02-07 17:29:18 | modified: 2020-02-07 17:40:02 | by: *bookofproofs* | references: [581]

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[581] **Forster Otto**: “Analysis 1, Differential- und Integralrechnung einer Veränderlichen”, Vieweg Studium, 1983