parmenides51 wrote: Find all nonnegative real numbers $a,b,c$ such that $$\frac{4a+9b+25c}{2a+3b+5c}+\frac{4b+9c+25a}{2b+3c+5a}+\frac{4c+9a+25b}{2c+3a+5b}=10$$ $\bullet x,y,z>0$ and $a,b,c \ge 0 \Longrightarrow \sum \frac{ax^2+by^2+cz^2}{ax+by+cz} \ge \sum \frac{ax+by+cz}{a+b+c}=x+y+z$ $\bullet \sum \frac{ax^2+by^2+cz^2}{ax+by+cz}=x+y+z \iff \left(a>0;b=0;c=0 \right) \vee \left(a=0;b>0;c=0 \right)\vee\left(a=0;b=0;c>0 \right).$ .