Hello, the papers aren't allowed to be published. Please detele this post. Edit: this was proposed by me(not created by me).

Not for this problem. Only problem 5 is from the shortlist. The solution is already public on : https://pregatirematematicaolimpiadejuniori.wordpress.com/2017/04/23/ro2017/

MathGan wrote: a) Find : $A=\{(a,b,c) \in \mathbb{R}^{3} | a+b+c=3 , (6a+b^2+c^2)(6b+c^2+a^2)(6c+a^2+b^2) \neq 0\}$ b) Prove that for any $(a,b,c) \in A$ next inequality hold : \begin{align*} \frac{a}{6a+b^2+c^2}+\frac{b}{6b+c^2+a^2}+\frac{c}{6c+a^2+b^2} \le \frac{3}{8} \end{align*} Easy C-S.

MathGan wrote: a) Find : $A=\{(a,b,c) \in \mathbb{R}^{3} | a+b+c=3 , (6a+b^2+c^2)(6b+c^2+a^2)(6c+a^2+b^2) \neq 0\}$ b) Prove that for any $(a,b,c) \in A$ next inequality hold : \begin{align*} \frac{a}{6a+b^2+c^2}+\frac{b}{6b+c^2+a^2}+\frac{c}{6c+a^2+b^2} \le \frac{3}{8} \end{align*} $$\iff\frac{b^2+c^2}{(c+a)^2+(a+b)^2}+\frac{c^2+a^2}{(a+b)^2+(b+c)^2}+\frac{a^2+b^2}{(b+c)^2+(c+a)^2}\ge \frac{3}{4}$$

Let $ a,b,c> 0 . $ Prove that$$ \frac{a^2}{(c+a)^2+(a+b)^2}+\frac{b^2}{(a+b)^2+(b+c)^2}+\frac{c^2}{(b+c)^2+(c+a)^2} \geq \frac{3} {8}$$

Let $ a,b,c>0 . $ Prove that $$\frac{a^3}{(a+b)^2+(a+c)^2}+\frac{b^3}{(b+c)^2+(b+a)^2}+\frac{c^3}{(c+a)^2+(c+b)^2} \geq \frac{3(a^2+b^2+c^2)}{8(a+b+c)}$$