Example 1.6.2. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). Find the minimum of the expression
\[3^{-a^{2}}+3^{-b^{2}}+3^{-c^{2}}\]
SOLUTION. We will again propose and solve the general problem: for each real number $k>0$, find the minimum of the following expression
$$P=k^{a^{2}}+k^{b^{2}}+k^{c^{2}}$$

Certainly, if $k \geq 1$ then $P \geq 3 k$ by AM-GM inequality. Therefore we only need to consider the remaining case $k \leq 1$. WLOG, assume that $a \geq b \geq c$. Let $t=\frac{a+b}{2}, u=$ $\frac{a-b}{2}$, then $t \geq 1$ and $a=t+u, b=t-u$. Let $k'=\frac{1}{k} \geq 1$ and consider the following function
$$g(u)=k^{(t-u)^{2}}+k^{(t+u)^{2}}+k^{c^{2}}$$

Since
$$g'(u)=2 \ln k \cdot(t+u) k^{(t+u)^{2}}-2 \ln k \cdot(t-u) k^{(t-u)^{2}}$$
we deduce that $g'(u)=0 \Leftrightarrow \ln (t+u)-\ln (t-u)=-4 t u \ln k$. Letting now
$$h(u)=\ln (t+u)-\ln (t-u)+4 t u \ln k$$
we infer that
$$h'(u)=\frac{1}{t+u}+\frac{1}{t-u}+4 \ln k \cdot t=\frac{2 t}{t^{2}-u^{2}}+4 t \ln k$$

Therefore $h'(u)=0 \Leftrightarrow 2(t^{2}-u^{2}) \ln k=-1 \Leftrightarrow 2 a b \ln k'=1$. Now we divide the problem into two smaller cases
(i) The first case. If $a b, b c, c a \leq \frac{1}{2 \ln k'}$, then
$$k^{a^{2}}+k^{b^{2}}+k^{c^{2}} \geq k^{9 / 4}+2 k^{\frac{1}{2 \ln k'}}=k^{9 / 4}+2 e^{-1 / 2}$$
(ii) The second case. If $a b \geq \frac{1}{2 \ln k'}$. From the previous result, we deduce that
$$h'(u)=0 \Leftrightarrow u=0 \Rightarrow g'(u)=0 \Leftrightarrow u=0$$
therefore
$$g(u) \geq g(0)=2 k^{t^{2}}+k^{(3-2 t)^{2}}=f(t)$$

Our remaining work is to find the minimum of $f(t)$ for $\frac{3}{2} \geq t \geq 1$. Since
$$f'(t)=4 \ln k\left(t \cdot k^{t^{2}}-(3-2 t) \cdot k^{(3-2 t)^{2}}\right)$$
we refer that $f'(t)=0 \Leftrightarrow(3-3 t)(3-t) \ln k'=\ln (3-2 t)-\ln t$. Denote
$$q(t)=(3-3 t)(3-t) \ln k'-\ln (3-2 t)-\ln t$$
then
$$q'(t)=(6 t-12) \ln k'+\frac{2}{3-2 t}+\frac{1}{t}=(6 t-12) \ln k'+\frac{3}{t(3-2 t)}$$

In the range $\left[1, \frac{3}{2}\right]$, the function $t(3-2 t)(2-t)=2 t^{3}-7 t^{2}+6 t$ is decreasing, hence the equation $q'(t)=0$ has no more than one real root. By Rolle's theorem, the equation $f'(t)$ has no more than two roots in $\left[1, \frac{3}{2}\right]$. It's then easy to get that
$$f(t) \geq \min \left(f(1), f\left(\frac{3}{2}\right)\right)=\min \left(3 k, 1+2 k^{9 / 4}\right)$$

According to the previous solution, the following inequality holds
$$k^{a^{2}}+k^{b^{2}}+k^{c^{2}} \geq \min \left(3 k, 1+2 k^{9 / 4}, 2 e^{-1 / 2}+k^{9 / 4}\right)$$

Notice that if $k \leq 1 / 3$ then
$$\min \left(3 k, 1+2 k^{9 / 4}, 2 e^{-1 / 2}+k^{9 / 4}\right)=3 k$$

Otherwise, if $k \geq 1 / 3$ then
$$\begin{array}{c}
1+2 k^{9 / 4} \leq 2 e^{-1 / 2}+k^{9 / 4} \\
\Rightarrow \min \left(3 k, 1+2 k^{9 / 4}, 2 e^{-1 / 2}+k^{9 / 4}\right)=\min \left(3 k, 1+2 k^{9 / 4}\right)
\end{array}$$

Therefore, we can conclude that
$$k^{a^{2}}+k^{b^{2}}+k^{c^{2}} \geq \min \left(3 k, 1+2 k^{9 / 4}, 2 e^{-1 / 2}+k^{9 / 4}\right)=\min \left(3 k, 1+2 k^{9 / 4}\right)$$

The initial problem is a special case for $k=\frac{1}{3}$. In this case, we have
$$3^{-a^{2}}+3^{-b^{2}}+3^{-c^{2}} \geq 1$$

However, if $k=\frac{1}{2}$ then the following stranger inequality holds
$$2^{-a^{2}}+2^{-b^{2}}+2^{-c^{2}} \geq 1+2^{-5 / 4}$$