Cyclic inequality How can we prove that:
$$a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10}+a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20}\geq 2(a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}), \ \forall\ a,b,c\geq 0.$$
I proved only that $32S_1+18S_2\geq 2 S$, using AM-GM inequality, where 
$S_1=a^{60} c^{10} +b^{60}a^{10}+c^{60}b^{10},$
$S_2=a^{50} c^{20} +b^{50}a^{20}+c^{50}b^{20},$
$S=a^{51}b^{9}c^{10}+b^{51}c^9 a^{10}+c^{51}a^9 b^{10}$
Note that Muirhead's inequality is not working there...
 A: Update 2019/10/05 (Simplify some proof.)
With computer, here is a solution:
Clearly, we only need to prove the case when $a, b, c > 0$.
Let $x=a^{10}, \ y = b^{10}, \ z=c^{10}$. The inequality is written as
$$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge
2x^5yz\sqrt[10]{\frac{x}{y}} + 2xy^5z\sqrt[10]{\frac{y}{z}} + 2xyz^5\sqrt[10]{\frac{z}{x}}.$$
We will use the following bound:
$$\sqrt[10]{u} \le f(u) = \frac{11 u^2 + 418 u + 171}{20(19 u + 11)}, \quad \forall u > 0$$ 
which follows from (letting $u = v^{10}$)
\begin{align}
&11 v^{20} + 418 v^{10} + 171 - 20(19v^{10}+11)v\\
=\ & (v-1)^4\\
&\times \left(11 v^{16}+44 v^{15}+110 v^{14}+220 v^{13}+385 v^{12}+616 v^{11}+924 v^{10}+1320 v^9+1815 v^8\right.\\
&\qquad \left. +2040 v^7+2044 v^6+1876 v^5+1585 v^4+1220 v^3+830 v^2+464 v+171\right).
\end{align}
With the bound above, it suffices to prove that
$$xy^6 + yz^6 + zx^6 + x^2y^5 + y^2z^5 + z^2 x^5 \ge
2x^5yzf(x/y) + 2xy^5zf(y/z) + 2xyz^5f(z/x)$$ or $F(x,y,z)\ge 0$
where $F(x,y,z)$ is a homogeneous polynomial of degree $10$.
We use the Buffalo Way. WLOG, assume that $z = \min(x,y,z) $. There are two possible cases:
1) $z \le y \le x$: Let $z=1, \ y=1+s, \ x = 1+s + t; \ s,t \ge 0$. Then, $F(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
2) $z \le x \le y$: Let $z=1, \ x=1+s, \ y=1+s+t; \ s,t\ge 0$. Then, $F(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
We are done.
