Prove that $\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$

(Homogenization)We need to prove that (for $a, b, c \ge 0$, at most one of them is zero) \begin{align} \frac{a+b+c}{3} \ge \sqrt[53]{\frac{a^4+b^4+c^4}{3}\Big(\frac{(a+b)(b+c)(c+a)}{8}\Big)^{49/3}}. \end{align} WLOG, assume that $c=1$ and $a+b \ge 1$. Let $p=a+b, \ q = ab$. Using $a^4+b^4 = p^4-4p^2q+2q^2$, taking logarithm on both sides, it suffices to prove that \begin{align} f(p, q) = 53\ln \frac{p+1}{3} - \ln \frac{p^4-4p^2q+2q^2+1}{3} - \frac{49}{3}\ln p - \frac{49}{3}\ln (1+p+q) + \frac{49}{3}\ln 8 \ge 0. \end{align} Since $q \le \frac{p^2}{4}$ and $p\ge 1$, we have $52p^2-3p-3-55q \ge 52p^2-3p-3-55(\frac{p^2}{4}) = \frac{153}{4}p^2-3p-3 > 0$. Thus, we have \begin{align} \frac{\partial f}{\partial q} &= -\frac{2(52p^2-3p-3-55q)^2 - 2713\, p^4 - 36\, p^3 - 54\, p^2 - 36\, p + 2677}{165(p^4-4p^2q+2q^2+1)(1+p+q)}\\ &\le -\frac{2(\frac{153}{4}p^2-3p-3)^2 - 2713\, p^4 - 36\, p^3 - 54\, p^2 - 36\, p + 2677}{165(p^4-4p^2q+2q^2+1)(1+p+q)}\\ &= -\frac{31p^4-72p^3-72p^2+392}{24(p^4-4p^2q+2q^2+1)(1+p+q)}\\ &\le 0. \end{align} Thus, we have $f(p, q) \ge f(p, \frac{p^2}{4})$. It suffices to prove that $f(p, \frac{p^2}{4})\ge 0$ or $$g(p) = 53\ln \frac{p+1}{3} - \ln \frac{p^4+8}{24} - \frac{49}{3}\ln p - \frac{49}{3}\ln (1+p+\frac{p^2}{4}) + \frac{49}{3}\ln 8\ge 0.$$ Note that $$g'(p) = \frac{(p-2)(37\, p^4 - 48\, p^3 - 96\, p^2 - 96\, p + 392)}{3(2+p)(p+1)(p^4+8)p}.$$ Since $37\, p^4 - 48\, p^3 - 96\, p^2 - 96\, p + 392 > 0$, we have $g'(p) < 0$ for $1\le p < 2$ and $g'(p) > 0$ for $2 < p$. Thus, $g(p) \ge g(2) = 0.$ We are done.