Too easy. After Ravi substitution $( a=x+y,b=y+z,c=z+x)$ we get $^3\sqrt{\prod \left( \sum x^2+\sum xy+2xy\right)}>$ $^3\sqrt{\left(\sum x^2+\sum xy \right)^3}=\sum x^2+\sum xy=\frac{a^2+b^2+c^2}{2}$

Let $a,b,c\geq 0$ and $ab+bc+ca\neq0. $ Prove that $$\frac{2}{3}(ab+bc+ca)\leq \sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)} \leq \frac{2}{3}(a^2+b^2+c^2)$$ Let $a,b,c$ be the side lengths of a triangle. Prove that $$\sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)-24(a-b)^24(b-c)^24(c-a)^2} > \frac{a^2+b^2+c^2}{2}. $$

But it doesn't right. When(a,b,c) ~(5,1,0), $$\sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)}=5\leq13=\frac{1}{2}(a^2+b^2+c^2)$$Unless $a,b,c$ be the side lengths of a triangle.

sqing wrote: Let $a,b,c\geq 0$ and $ab+bc+ca\neq0. $ Prove that $$\frac{1}{2}(a^2+b^2+c^2)\leq\sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)} \leq \frac{2}{3}(a^2+b^2+c^2)$$ $$The\; right-mid=\frac{27\sum_{cyc}a^4(b-c)^2+21\sum_{cyc}a^2b^2(a-b)^2+6\sum_{cyc}(a^3-b^3)^2+4(a^6+b^6+c^6-3a^2b^2c^2)}{2}\geq 0$$

How to understand the official solution?

zhaoli wrote: How to understand the official solution? Where does the official answer come from? Thanks.

rafaello wrote: Let $a$, $b$, $c$ be the side lengths of a triangle. Prove that $$ \sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)} > \frac{a^2+b^2+c^2}{2}. $$ $$a^2+bc>\frac{a^2+(b-c)^2}{2}+bc=\frac{a^2+b^2+c^2}{2},...$$7615014650

I think it's here Locked document but you need the appropriate credentials to access it. It's taken from this link: Baltic Way 2021 website

Another

WLOG, let $b,c \geq a$. Using Holder's, we have $$\sqrt[3]{(a^2+bc)(b^2+ca)(c^2+ab)}=\sqrt[3]{(bc+a^2)(b^2+ca)(c^2+ab)} \geq bc+a\sqrt[3]{abc} $$so it suffices to prove that $2bc+2a\sqrt[3]{abc}>a^2+b^2+c^2$, which rearranges to $2a\sqrt[3]{abc} >a^2+(b-c)^2$. Since $b,c \geq a$, we have $2a\sqrt[3]{abc} \geq 2a^2=a^2+a^2 \geq a^2+(b-c)^2$, as desired.

a=y+z,b=x+z,c=x+y very easy problem

By Ravi Transformation we can rewrite $a = x+y$,$b = y+z$ and $c = z+x$ then we have to prove $\sqrt[3]{( \sum x^2+\sum xy+2xy)( \sum x^2+\sum xy+2yz)( \sum x^2+\sum xy+2zx)} > \sum x^2+\sum xy$ which is obviously true cause $\sqrt[3]{( \sum x^2+\sum xy+2xy)( \sum x^2+\sum xy+2yz)( \sum x^2+\sum xy+2zx)} > \sqrt[3]{( \sum x^2+\sum xy)^3} = \sum x^2+\sum xy$.