Min of $\frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $ I got this problem I tried several time to solve it by many inequalities but I got stuk. My question is how I get the minimum value of  $$ \frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $$ if you know that  $a, b, c \in (0, \infty) $ and $ a+b+c=1$?  Any hint? 
 A: First note that 
$$(a^3-b^3)(a^7-b^7)\ge 0\tag{1}$$
so
$$2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3).\tag{2}$$
It follows that
$$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{c^{10}+a^{10}}{c^7+a^7}\ge a^3+b^3+c^3.$$
Note that the equality happens when $a=b=c$. Next, we show that $a^3+b^3+c^3$ attains its minimum also when $a=b=c$. This can be seen by showing
$$3(a^3+b^3+c^3)\ge (a+b+c)(a^2+b^2+c^2)\ge (a+b+c)\frac{(a+b+c)^2}{3}\tag{3},$$
using Rearrangement Inequality, which can be proved as in (1) $\implies$ (2).
Or we can use Cauchy-Schwarz inequality
$$(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2\ge \frac{(a+b+c)^4}{3^2}.$$
A: Another (though similar) way...  by Chebyshev's inequality applied twice:
$$ \sum_{cyc} \frac{a^{10}+b^{10}}{a^7+b^7} \ge \sum_{cyc}\frac{a^3+b^3}2 = \sum_{cyc}a^3\ge \frac19(a+b+c)^3 =\frac19$$
with equality when $a=b=c=\frac13$, so that's the min.
A: $2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3)$ with chebishev inequality, then $$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{a^{10}+c^{10}}{a^7+c^7} \ge a^3+b^3+c^3 \ge 3 (\frac{a+b+c}{3})^3=\frac{1}{9}$$ because with power mean we have $ (\frac{a^3+b^3+c^3}{3})^\frac{1}{3} \ge (\frac{a+b+c}{3})$ 
