55. Find the best constant $k$ (minimum value) such that the following inequality holds for non-negative real numbers $a, b, c$ satisfying $a+b+c=3$.
$$a^{k}+b^{k}+c^{k} \geq a b+b c+c a$$
(Genralization Of Russia M0 2000)
Solution: In Example 1.1.1 of Chapter 1, the inequality has already been proven when $k=\frac{1}{2}$. Therefore, for $k \geq \frac{1}{2}$, it holds. Now consider the case when $k \leq \frac{1}{2}$.

Lemma: Let $a, b \geq 0, a+b=2 t \geq 1$, then we have $a^{k}+b^{k}-a b \geq \min \left\{(2 t)^{k}, 2 t^{k}-t^{2}\right\}$.

In fact, without loss of generality, assume $a \geq b$, then there exists a non-negative real number $x$ such that $a=t+x, b=t-x$. Consider the function
$$f(x)=(t+x)^{k}+(t-x)^{k}-t^{2}+x^{2}$$

Then
$$\begin{array}{l}
f^{\prime}(x)=k(t+x)^{k-1}-k(t-x)^{k-1}+2 x \\
f^{\prime \prime}(x)=k(k-1)\left[(t+x)^{k-2}+(t-x)^{k-2}+2\right] \\
f^{\prime \prime}(x)=k(k-1)(k-2)\left[(t+x)^{k-3}-(t-x)^{k-3}\right]
\end{array}$$

So $f^{\prime \prime \prime}(x) \leq 0$, therefore $f^{\prime \prime}(x)$ is a monotonic function, so $f^{\prime}(x)$ has no more than two roots.
Since $f^{\prime}(0)=0$ and
$$f^{\prime \prime}(0)=2 k(k-1) t^{k-2}+2=2-2 k(1-k) \geq 0$$

Thus, $f(x)$ reaches its minimum value only at $x=0$ or $x=t$.
Returning to our problem, without loss of generality, assume $a \geq b \geq c$, and set $a+b=2 t \geq 1$, then
$$a^{k}+b^{k}+c^{k}-(a b+b c+c a) \geq \min \left\{(2 t)^{k}, 2 t^{k}-t^{2}\right\}-2 c t+c^{k}$$
(i) If $(2 t)^{k} \leq 2 t^{k}-t^{2}$, applying the lemma to $2 t, c$, we get
$$a^{k}+b^{k}+c^{k}-(a b+b c+c a) \geq(2 t)^{k}+c^{k}-c \cdot 2 t \geq \min \left\{(2 t+c)^{k}, 2\left(t+\frac{c}{2}\right)^{k}-\left(t+\frac{c}{2}\right)^{2}\right\}$$

Since $2 t+c=3$, we have
$$a^{k}+b^{k}+c^{k}-(a b+b c+c a) \geq \min \left\{3^{k}, 2 \cdot \frac{3^{k}}{2^{k}}-\frac{9}{4}\right\}$$
(ii) If $(2 t)^{k} \geq 2 t^{k}-t^{2}$. We will prove that $g(t) \geq 0$. Where
$$g(t)=2 t^{k}+(3-2 t)^{k}-2 t(3-2 t)+t^{2}=2 t^{k}+(3-2 t)^{k}-6 t+3 t^{2}$$

Notice that
$$\begin{array}{l}
g^{\prime}(t)=2 k t^{k-1}-2 k(3-2 t)^{k-1}-6+6 t \\
g^{\prime \prime}(t)=2 k(k-1)\left[t^{k-2}-2(3-2 t)^{k-2}\right]+6 \\
g^{\prime \prime \prime}(t)=2 k(k-1)(k-2)\left[t^{k-3}-4(3-2 t)^{k-3}\right]
\end{array}$$

Since $g^{\prime \prime \prime}(t)$ has no real roots when $t \geq 1$, we have that $g^{\prime}(t)$ has no more than two real roots. Thus, we have
$$\min _{1 \leq t \leq \frac{3}{2}} g(t)=\min \left(g(1), g\left(\frac{3}{2}\right)\right)=\min \left(0,2 \cdot \frac{3^{k}}{2^{k}}-\frac{9}{4}\right)$$

Based on these results, we have, for all positive real numbers $k$,
$$a^{k}+b^{k}+c^{k}-(a b+b c+c a) \geq \min \left(0,2 \cdot \frac{3^{k}}{2^{k}}-\frac{9}{4}\right)$$

Therefore, the best constant $k$ is
$$2 \cdot \frac{3^{k}}{2^{k}}=\frac{9}{4} \Leftrightarrow k=\frac{2 \ln 3-3 \ln 2}{\ln 3-\ln 2} \approx 0.2905$$

In this case, equality holds when $a=b=c=1$ or $a=b=\frac{3}{0}, c=0$ and their permutations.