For every positive real numbers $a,b,c$ for which $a+b+c=3$ we have: $$a^5+b^5+c^5+8(ab+bc+ca)\ge 27.$$
My ideas is:
We can apply Chebyshev inequality $$\sum a^5 =\sum a\cdot a^4\ge \frac13 \left(\sum a\right)\left(\sum a^4\right)=\sum a^4.$$ Remains to show that $\sum a^4+8\sum ab \ge 27.$
As $2(ab+bc+ca)=9-a^2-b^2-c^2$ we have to prove $$a^5+b^5+c^5-4(a^2+b^2+c^2)\ge -9$$ Now $$x^5-4x^2\ge -3x\iff x (x - 1)^2 (x^2 + 2 x + 3)\ge 0 \;\forall x\ge 0$$ so $$ a^5+b^5+c^5-4(a^2+b^2+c^2)\ge -3(a+b+c)=-9$$ Done