Given $a,b,c$. Prove inequality

Question

Given $a,b,c$ are non-negative number. Prove that $$3(a^2+b^2+c^2)\ge\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+(a-b)^2+(b-c)^2+(c-a)^2\ge(a+b+c)^2$$

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By C-S: $3(a^2+b^2+c^2)\ge(a+b+c)^2$

And prove $3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2$

Because $\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0$

Hence $3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)$

$\Rightarrow 3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2$ (by C-s)

Also prove $\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge\left(a+b+c\right)^2\left(1\right) (*)$

Help me prove (*)

Answer 1

We need to prove that $$(a^2+b^2+c^2)(ab+ac+bc)+\sum_{cyc}(a^2-b^2)^2\geq(a^2+b^2+c^2)^2$$ or $$\sum_{cyc}(a^4+a^3b+a^3c-4a^2b^2+a^2bc)\geq0$$ or $$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+2\sum_{cyc}ab(a-b)^2\geq0,$$ which is true by Schur.

Done!

Answer 2

Setting $$a=x^2,b=y^2,c=z^2$$ then we have to prove that $$3 \left(x^4+y^4+z^4\right)-(x y+x z+y z) \left(x^2+y^2+z^2\right)-\left(x^2-y^2\right)^2-\left(z^2-x^2\right)^2-\left(y^2-z^2\right)^2\geq 0$$ and this is equivalent to $$\left(x^2+y^2+z^2\right) \left(x^2-x y-x z+y^2-y z+z^2\right)\geq 0$$ which is true. The right-hand side is equivalent to $$-\left(x^2+y^2+z^2\right)^2+(x y+x z+y z) \left(x^2+y^2+z^2\right)+\left(x^2-y^2\right)^2+\left(z^2-x^2\right)^2+\left(y^2-z^2\right)^2\geq 0$$ Setting $$y=x+u,z=x+u+v$$ we get $$x^2 \left(5 u^2+5 u v+5 v^2\right)+5 u^2 v^2+x \left(4 u^3+6 u^2 v+14 u v^2+6 v^3\right)+5 u v^3+v^4\geq 0$$ which is true.