Problem 55. Find the best constant $k$ (smallest) for the inequality
$$a^{k}+b^{k}+c^{k} \geq a b+b c+c a$$
to hold for $a, b, c$ are three non-negative real numbers with $a+b+c=3$.
SOLUTION. In example 1.1.1 in chapter I, this inequality is proved for \( k = \frac{1}{2} \) and therefore it's true for every \( k \geq \frac{1}{2} \). Consider the inequality in the case \( k \leq \frac{1}{2} \).

Lemma. Suppose \( a, b \geq 0 \) and \( a + b = 2t \geq 1 \) then we have
\[ a^k + b^k - ab \geq \min \left((2t)^k, 2t^k - t^2\right) \]

Indeed, WLOG assume that \( a \geq b \). There exists a non-negative real number \( x \) with \( a = t + x, b = t - x \). Consider the function
\[ f(x) = (t + x)^k + (t - x)^k - t^2 + x^2 \]
then
\[ \begin{aligned}
f'(x) & = k(t + x)^{k-1} - k(t - x)^{k-1} + 2x \\
f''(x) & = k(k-1)\left((t + x)^{k-2} + (t - x)^{k-2}\right) + 2 \\
f'''(x) & = k(k-1)(k-2)\left((t + x)^{k-3} - (t - x)^{k-3}\right)
\end{aligned} \]

So \( f'''(x) \leq 0 \), hence \( f''(x) \) is a monotone function and therefore \( f'(x) \) has no more than two roots. Since \( f'(0) = 0 \) and
\[ f''(0) = 2k(k-1)t^{k-2} + 2 = 2 - 2k(1 - k) \geq 0 \]
we conclude that \( f \) attains the minimum at \( x = 0 \) or \( x = t \) only.
Returning to our problem, WLOG, assume that \( a \geq b \geq c \). Let \( a + b = 2t \geq 1 \), then
\[ a^k + b^k + c^k - (ab + bc + ca) \geq \min \left\{(2t)^k, 2t^k - t^2\right\} - 2ct + c^k \]
(i). \( (2t)^k \leq 2t^k - t^2 \). Using the lemma for \( 2t \) and \( c \), we obtain
\[ a^k + b^k + c^k - (ab + bc + ca) \geq (2t)^k + c^k - c \cdot 2t \geq \min \left\{(2t + c)^k, 2\left(t + \frac{c}{2}\right)^k - \left(t + \frac{c}{2}\right)^2\right\} \]

Since \( 2t + c = 3 \), we can conclude that
\[ a^k + b^k + c^k - (ab + bc + ca) \geq \min \left\{3^k, 2 \cdot \frac{3^k}{2^k} - \frac{9}{4}\right\} \]
(ii). \( (2t)^k \geq 2t^k - t^2 \). We will prove that \( g(t) \geq 0 \) where
\[ g(t) = 2t^k + (3 - 2t)^k - 2t(3 - 2t) + t^2 = 2t^k + (3 - 2t)^k - 6t + 3t^2 \]

Notice that
\[ \begin{aligned}
g'(t) & = 2kt^{k-1} - 2k(3 - 2t)^{k-1} - 6 + 6t \\
g''(t) & = 2k(k-1)\left(t^{k-2} - 2(3 - 2t)^{k-2}\right) + 6 \\
g'''(t) & = 2k(k-1)(k-2)\left(t^{k-3} - 4(3 - 2t)^{k-3}\right)
\end{aligned} \]

Because \( g'''(t) \) has no roots if \( (t \geq 1) \), we infer \( g'(t) \) has no more than two roots. We deduce that
\[ \min_{1 \leq t \leq 3/2} g(t) = \min \left(g(1), g\left(\frac{3}{2}\right)\right) = \min \left(0, 2 \cdot \frac{3^k}{2^k} - \frac{9}{4}\right) \]

According to these results, we conclude that for all positive real \( k \)
\[ a^k + b^k + c^k - (ab + bc + ca) \geq \min \left(0, 2 \cdot \frac{3^k}{2^k} - \frac{9}{4}\right) \]

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 for \( a = b = c = 1 \) or \( a = b = \frac{3}{2}, c = 0 \) or permutations.