Exercise 8. Let $a \in \mathbb{R}$. What are the functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}$ such that, for all $x, y, z \in \mathbb{R}^{*}$, we have:

$$
\operatorname{af}\left(\frac{x}{y}\right)+\operatorname{af}\left(\frac{x}{z}\right)-f(x) f\left(\frac{y+z}{2}\right) \geqslant a^{2} ?
$$
Solution to Exercise 8: We proceed by analysis-synthesis. Suppose that $f$ is a solution to this equation, what must it satisfy?

The equation-statement must be true for all $x, y, z$, so it must be true in particular if $x=y=z$, which gives the equation (1):

$$
\operatorname{af}(1)+\operatorname{af}(1)-f(x) f(x) \geqslant a^{2}, \text{ for all } x \in \mathbb{R}^{*}
$$

In particular, for $x=1$, it follows that $0 \geqslant(f(1)-a)^{2}$, hence $f(1)=a$.

This result allows us to better understand equation (1), which becomes:

$$
2 a^{2}-f(x)^{2} \geqslant a^{2}, \text{ for all } x \in \mathbb{R}^{*}
$$

Thus, we have equation (2): $a^{2} \geqslant f(x)^{2}$ for all $x \in \mathbb{R}^{*}$. (3):

Now let's return to the equation-statement and set only $y=z=1$, to obtain the equation

$$
\operatorname{af}(x) \geqslant a^{2}, \text{ for all } x \in \mathbb{R}^{*}
$$

If $a=0$, equation (2) gives $f(x)=0=a$ for all $x \in \mathbb{R}^{*}$.

If $a>0$, equation (3) gives $f(x) \geqslant a>0$ and then (2) specifies $a \geqslant f(x)$, hence $f(x)=a$ for all $x \in \mathbb{R}^{*}$.

Finally, if $a<0$, equation (3) gives $f(x) \leqslant a<0$ and then (2) specifies $a \leqslant f(x)$, hence $f(x)=a$ for all $x \in \mathbb{R}^{*}$.

In short, in all cases, if $f$ is a solution, then $f$ is constant and equal to $a$ for all $x \in \mathbb{R}^{*}$.

And conversely, the constant function equal to $a$ being a solution, we have indeed found exactly the solutions to the functional equation.