Problem. Let $a,b,c$ be real numbers. Find minimum and maximum$$P=\frac{bc}{a^2+2b^2+2c^2}+\frac{ca}{b^2+2a^2+2c^2}+\frac{ab}{c^2+2b^2+2a^2}.$$
I worked on maximum number of days but I did not find any interesting idea. The minimum is out of my reach.
Set $a=b$ we will prove $P\le \dfrac{3}{5}.$ The big trouble here is real variables.
I think we should split $P$ into cases: three of them be positive, negative,...I am still stuck here.
Hope to see some ideas. Thank you.
On
For a maximal value SOS helps.
Indeed, for $a=b=c=1$ we obtain a value $\frac{3}{5}.$
We'll prove that it's a maximal value.
We need to prove that: $$\sum_{cyc}\frac{ab}{2a^2+2b^2+c^2}\leq\frac{3}{5}$$ or $$\sum_{cyc}\left(\frac{1}{5}-\frac{ab}{2a^2+2b^2+c^2}\right)\geq0$$ or $$\sum_{cyc}\frac{2a^2+2b^2-5ab+c^2}{2a^2+2b^2+c^2}\geq0$$ or $$\sum_{cyc}\frac{(c-a)(5b-4a+c)-(b-c)(4a-4b+c)}{2a^2+2b^2+c^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{5c-4b+a}{2b^2+2c^2+a^2}-\frac{5c-4a+b}{2a^2+2c^2+b^2}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(2a^2+2b^2+c^2)(10c^2+5(a+b)c+6a^2-3ab+6b^2)\geq0,$$ which is true because $$10c^2+5(a+b)c+6a^2-3ab+6b^2=10\left(c+\frac{a+b}{4}\right)^2-\frac{5(a+b)^2}{8}+6a^2-3ab+6b^2\geq$$ $$\geq\frac{43a^2-34ab+43b^2}{8}\geq0.$$ It seems that the minimal value is equal to $-\frac{1}{4}$ and occurs for $(a,b,c)=(1,-1,0)$,
but I have no a proof for this statement.
For a minimal value, SOS helps well too.
I give a proof assisted by computer. After full expanding, we get equivalent SOS expression $$\frac{2 \left(73267 x^2-15728 x y-15728 x z+73267 y^2-15728 y z+73267 z^2\right) (x+y+z)^4}{79475}+\\ \frac{\left(252 x^2 y+252 x^2 z+759 x y^2+413 x y z+759 x z^2+507 y^3-519 y^2 z-519 y z^2+507 z^3\right)^2}{238425}+\\ \frac{\left(507 x^3+759 x^2 y-519 x^2 z+252 x y^2+413 x y z-519 x z^2+252 y^2 z+759 y z^2+507 z^3\right)^2}{238425}+\\ \frac{\left(507 x^3-519 x^2 y+759 x^2 z-519 x y^2+413 x y z+252 x z^2+507 y^3+759 y^2 z+252 y z^2\right)^2}{238425}\ge 0.$$Equality case: $(0,t,-t)$ where $t\neq 0.$