Prove that inequality $6(\sum_{cyc}ab)+\sum_{cyc}a(b-c)^2$

Question

For $a,b,c>0$ satisfy $a+b+c=1$. Prove that $$6(ab+bc+ca)+a(b-c)^2+b(a-c)^2+c(a-b)^2\le2$$

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We have: $6(ab+bc+ac)+1\le 3(a+b+c)^2$

$\Leftrightarrow a^2+b^2+c^2\ge \frac{1}{3}$ (right by C-S with a+b+c=1)

Hence $LHS=6(ab+bc+ca)+1+a(b-c)^2+b(a-c)^2+c(a-b)^2\le3$

$a(b-c)^2+b(a-c)^2+c(a-b)^2\le3(a+b+c)^2=3 $

The last inequality right if $a(b-c)^2+b(a-c)^2+c(a-b)^2\le 0$ but $a(b-c)^2+b(a-c)^2+c(a-b)^2\ge0$ so wrong. Help

Answer 1

We need to prove that $$6(a+b+c)(ab+ac+bc)+\sum_{cyc}(a^2b+a^2c-2abc)\leq2(a+b+c)^3$$ or $$6\sum_{cyc}(a^2b+a^2c+abc)+\sum_{cyc}(a^2b+a^2c-2abc)\leq2\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)$$ or $$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or $$\sum_{cyc}(a^3-a^2b-ab^2+b^3)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)\geq0.$$

Done!

Answer 2

$$p=a+b+c =1\ ,\ q=ab+bc+ca\ ,\ r=abc$$

$$\Leftrightarrow 7q-9r\le 2$$

by Shur : $ 1+9r\ge 4q \Rightarrow 9r \ge 4q-1 \ge 7q-2 \Leftrightarrow 1 \ge 3q$