Problem 1136

For example, consider the function $\sqrt{x y}$ defined for positive values. Forming the difference between the two sides of the symmetric Jensen inequality for this function:

$$\begin{aligned} & \sqrt{\left(\frac{x_{1}+x_{2}}{2}\right)\left(\frac{y_{1}+y_{2}}{2}\right)}-\frac{\sqrt{x_{1} y_{1}}+\sqrt{x_{2} y_{2}}}{2}= \\ = & \frac{1}{2} \cdot \frac{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(\sqrt{x_{1} y_{1}}+\sqrt{x_{2} y_{2}}\right)^{2}}{\sqrt{\left(x_{1}+y_{2}\right)\left(y_{1}+y_{2}\right)}+\sqrt{x_{1} y_{1}}+\sqrt{x_{2} y_{2}}} \end{aligned}$$

Here, the denominator is positive, and the numerator can be further transformed by expanding the square terms and simplifying:

$$x_{1} y_{2}+x_{2} y_{1}-2 \sqrt{x_{1} y_{1} x_{2} y_{2}}$$

Here, the first two terms are the arithmetic mean of $x_{1} y_{2}$ and $x_{2} y_{1}$, and the subtrahend is twice their geometric mean. Thus, the difference cannot be negative. The expression can be zero, however, if $x_{1} y_{2}=x_{2} y_{1}$, i.e., $x_{1} / y_{1}=x_{2} / y_{2}$. Therefore, the surface is concave in a broader sense. On the $(x, y)$ plane, straight lines passing through the initial point also draw straight lines on the surface.

For the function $\stackrel{a}{\log }\left(a^{x}+a^{y}\right)$

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according to the previously proven inequality, assuming that $a>1$. Equality can hold if $a^{x_{1}-y_{1}}=a^{x_{2}-y_{2}}$. If $0<a<1$, the function is convex if $a>1$ and concave in a broader sense if $0<a<1$.

Now let's examine the function $x^{q_{1}} x^{q_{2}}$ where $x, y, q_{1}$, and $q_{2}$ are positive and $q_{1}+q_{2}=1$. Here, we need to compare the expressions

$$\frac{x_{1}^{q_{1}} y_{2}^{q_{2}}+x_{2}^{q_{1}} y_{2}^{q_{2}}}{2} \text { and }\left(\frac{x_{1}+x_{2}}{2}\right)^{q_{1}}\left(\frac{y_{1}+y_{2}}{2}\right)^{q_{2}}=\frac{1}{2}\left(x_{1}+x_{2}\right)^{q_{1}}\left(y_{1}+y_{2}\right)^{q}$$

We can transform their ratio using the fact that the weighted geometric mean of two quantities is less than or equal to their weighted arithmetic mean.

$$\begin{gathered} \frac{x_{1}^{q_{1}} y_{1}^{q_{2}}+x_{2}^{q_{1}} y_{2}^{q_{2}}}{\left(x_{1}+x_{2}\right)^{q_{1}}\left(y_{1}+y_{2}\right)^{q_{2}}}=\left(\frac{x_{1}}{x_{1}+x_{2}}\right)^{q_{1}}\left(\frac{y_{1}}{y_{1}+y_{2}}\right)^{q_{2}}+\left(\frac{x_{2}}{x_{1}+x_{2}}\right)^{q_{1}}\left(\frac{y_{2}}{y_{1}+y_{2}}\right)^{q_{2}} \leq \\ \leq q_{1} \frac{x_{1}}{x_{1}+x_{2}}+q_{2} \frac{y_{1}}{y_{1}+y_{2}}+q_{1} \frac{x_{2}}{x_{1}+x_{2}}+q_{2} \frac{y_{2}}{y_{1}+y_{2}}=q_{1}+q_{2}=1 \end{gathered}$$

Equality can hold if

$$\frac{x_{1}}{x_{1}+x_{2}}=\frac{y_{1}}{y_{1}+y_{2}} \quad \text { and } \quad \frac{x_{2}}{x_{1}+x_{2}}=\frac{y_{2}}{y_{1}+y_{2}}$$

i.e., if $x_{1} / y_{1}=x_{2} / y_{2}$. Therefore, the function $x^{q_{1}} y^{q_{2}}$ is concave in a broader sense if $q_{1}+q_{2}=1$. Similarly, for the function $x_{1}^{q_{1}} x_{2}^{q_{2}} \ldots x_{n}^{q_{n}}$ where $x_{1}, x_{2}, \ldots, x_{n} ; y_{1}, y_{2}, \ldots, y_{n} ; q_{1}, q_{2}, \ldots, q_{n}$ are positive and $q_{1}+q_{2}+\ldots+q_{n}=1$, we have

$$\begin{aligned} & \frac{x_{1}^{q_{1}} x_{2}^{q_{2}} \ldots x_{n}^{q_{n}}+y^{q_{1}} y^{q_{2}} \ldots y_{n}^{q_{n}}}{2} \\ & \left(\frac{x_{1}+y_{1}}{2}\right)^{q_{1}}\left(\frac{x_{2}+y_{2}}{2}\right)^{q_{2}} \ldots\left(\frac{x_{n}+y_{n}}{2}\right)^{q_{n}}= \\ & =\left(\frac{x_{1}}{x_{1}+y_{1}}\right)^{q_{1}}\left(\frac{x_{2}}{x_{2}+y_{2}}\right)^{q_{2}} \ldots\left(\frac{x_{n}}{x_{n}+y_{n}}\right)^{q_{n}}+\left(\frac{y_{1}}{x_{1}+y_{1}}\right)^{q_{1}}\left(\frac{y_{2}}{x_{2}+y_{2}}\right)^{q_{2}} \ldots\left(\frac{y_{n}}{y_{n}+y_{n}}\right)^{q_{n}} \leq \\ & \leq q_{1} \frac{x_{1}}{x_{1}+y_{1}}+q_{2} \frac{x_{2}}{x_{2}+y_{2}}+\ldots+q_{n} \frac{x_{n}}{x_{n}+y_{n}}+q_{1} \frac{y_{1}}{x_{1}+y_{1}}+q_{2} \frac{y_{2}}{x_{2}+y_{2}}+\ldots+q_{n} \frac{y_{n}}{x_{n}+y_{n}}= \\ & =q_{1}+q_{2}+\ldots+q_{n}=1 \end{aligned}$$

Thus, the function is again concave in a broader sense. Let's write down the symmetric inequality for $k$ terms. Denote $k$ sets of $n$ numbers as $\left(x_{1}^{(1)}, x_{2}^{(1)}, \ldots, x_{n}^{(1)}\right),\left(x_{1}^{(2)}, x_{2}^{(2)}, \ldots, x_{n}^{(2)}\right), \ldots,\left(x_{1}^{(k)}, x_{2}^{(k)}, \ldots, x_{n}^{(k)}\right)$, then

$$\begin{gathered} \left(\frac{x_{1}^{(1)}+x_{1}^{(2)}+\ldots+x_{1}^{(k)}}{k}\right)^{q_{1}}\left(\frac{x_{2}^{(1)}+x_{2}^{(2)}+\ldots+x_{2}^{(k)}}{k}\right)^{q_{2}} \ldots\left(\frac{x_{n}^{(1)}+\ldots+x_{n}^{(k)}}{k}\right)^{q_{n}} \geq \\ \geq \frac{x_{1}^{(1) q_{1}} x_{2}^{(1) q_{2}} \ldots x_{n}^{(1) q_{n}}+x_{1}^{(2) q_{1}} \ldots x_{n}^{(2) q_{n}}+\ldots+x_{1}^{(k) q_{1}} x_{2}^{(k) q_{2}} \ldots x_{n}^{(k) q_{n}}}{k} . \end{gathered}$$

Here, since $q_{1}+q_{2}+\ldots q_{n}=1$, the denominators can be omitted.