Inequality about Schur

Question

Let a,b,c be the sides of triangle such that $a+b+c=1$. Prove that $$5(ab+bc+ca)\geq18abc+a+b+c$$ I tried to prove: $$5(ab+bc+ca)\geq18abc+a+b+c$$$$10(ab+ac+bc)\geq36abc+2(a+b+c)$$$$a(5b+5c-2-12bc)+b(5c+5a-2-12ca)+c(5a+5b-2-12ab)\geq0.$$I tried to prove $$5b+5c-2-12bc\geq0.$$I know $$-bc\geq-\frac{(b+c)^2}{4}.$$I want to prove $$5(b+c)-2-3(b+c)^2\geq0$$$$5(1-a)-2-3(1-a)^2\geq0$$and I got $$a(1-3a)\geq0$$which can be wrong.

Answer 1

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that: $$5(a+b+c)(ab+ac+bc)\geq18abc+(a+b+c)^3$$ or $$10(x+y+z)\sum_{cyc}(x^2+3xy)\geq8(x+y+z)^3+18\prod_{cyc}(x+y)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is indeed Schur.