Example 1.6.1. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). Find the maximum of the following expressions
(a) \(S_{2}=2^{a b}+2^{b c}+2^{c a}\).
(b) \(\quad S_{4}=4^{a b}+4^{b c}+4^{c a}\).
SOLUTION. Don't hurry to conclude that $\max S_{2}=6$ and $\max S_{4}=12$ because the reality is different. We figure out a solution by the mixing variable method and solve a general problem that involves both $(a)$ and $(b)$. WLOG, assume that $a \geq b \geq c$ and $k \geq 1$ is a positive real constant. Consider the following expression
$$f(a, b, c)=k^{a b}+k^{b c}+k^{c a}$$

Let $t=\frac{a+b}{2}, u=\frac{a-b}{2}$, then $t \geq 1, a=t+u, b=t-u$ and
$$f(a, b, c)=k^{t^{2}-u^{2}}+k^{c(t-u)}+k^{c(t+u)}=g(u)$$

Its derivative is
$$g^{\prime}(u)=2 u \ln k \cdot k^{t^{2}-u^{2}}+\ln k \cdot c k^{c t}\left(k^{c u}-k^{-c u}\right)$$

By Lagrange theorem, there exists a real number $r \in[-u, u]$ such that
$$\frac{k^{c u}-k^{-c u}}{2 u}=c k^{c r}$$
and therefore $k^{c u}-k^{-c u} \leq 2 u c k^{c u}$ (because $\left.r \leq u\right)$. Moreover, $c \leq 1$ and $c(t+u) \leq$ $(t-u)(t+u)=t^{2}-u^{2}$ so we obtain $g^{\prime}(u) \leq 0$. Thus
$$g(u) \leq g(0)=k^{t^{2}}+2 k^{c t}=k^{t^{2}}+2 k^{t(3-2 t)}=h(t)$$

We will prove that $h(t) \leq \max \left(h\left(\frac{3}{2}\right), h(1)\right) \forall k \geq 1$. Since
$$h^{\prime}(t)=2 t \ln k \cdot k^{t^{2}}+2(3-4 t) \ln k \cdot k^{t(3-2 t)}$$
we infer $h^{\prime}(t)=0 \Leftrightarrow 4 t-3=t \cdot k^{3 t(t-1)}$. Consider the following function
$$q(t)=3 t(t-1) \ln k-\ln (4 t-3)+\ln t$$

Because $q^{\prime}(t)=(6 t-3) \ln k-\frac{3}{t(4 t-3)}$ is a decreasing function of $t$ (when $t \geq 1$ ), we deduce that the equation $q^{\prime}(t)=0$ has no more than one root $t \geq 1$. According to Rolle's theorem, the equation $h^{\prime}(t)=0$ has no more than two roots $t \in\left[1, \frac{3}{2}\right]$. Moreover, because $h^{\prime}(1)=0$, we conclude that
$$h(t) \leq \max \left(h(1), h\left(\frac{3}{2}\right)\right)$$

According to this proof, we can synthesize a general result as follows
Let $a, b, c$ be non-negative real numbers with sum 3 . For all $k \geq 1$, we have
$$k^{a b}+k^{b c}+k^{c a} \leq \max \left(3 k, k^{9 / 4}+2\right)$$