Geometric inequality in triangle with inradius

Question

I am trying to examine if the inequality $ab+bc+ca \ge 12Rr+ \frac{a^2+b^2+c^2}{3}$ holds for a triangle. Since $ab+ca+ca \ge 4r(5R-r)$ (Bottema, p. 53) it is enough to prove $12r(2R-r) \ge a^2+b^2+c^2.$ I tried to use some other inequalities from Bottema but without any success. Any help?

Answer 1

In the standard notation we need to prove that: $$ab+ac+bc\geq12\cdot\frac{abc}{4S}\cdot\frac{2S}{a+b+c}+\frac{a^2+b^2+c^2}{3}$$ or $$3(a+b+c)(ab+ac+bc)\geq18abc+(a+b+c)(a^2+b^2+c^2)$$ or $$\sum_{cyc}(-a^3+2a^2b+2a^2c-3abc)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, since $$3b-a-c>2b+a-c-a-c=2(b-c)\geq0,$$ we obtain: $$\sum_{cyc}(-a^3+2a^2b+2a^2c-3abc)=$$ $$=\frac{1}{2}\sum_{cyc}(4a^2b+4ab^2-8abc-2a^3+2abc)=$$ $$=\frac{1}{2}\sum_{cyc}(4ab(b-c-(c-a))-(a+b+c)(a-b)^2)=$$ $$=\frac{1}{2}\sum_{cyc}(4(a-b)(ac-bc)-(a+b+c)(a-b)^2)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(3c-a-b)\geq$$ $$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-c)^2(3b-a-c))\geq$$ $$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-b)^2(3b-a-c))=$$ $$=(a-b)^2(b+c-a)\geq0.$$