Find the smallest real $k>0$ so that for every $x_1,x_2,x_3 \in [0,\frac{1}{2}]$ and $y_1,y_2,y_3 \in (0,\infty)$ with the condition $x_1+x_2+x_3=y_1+y_2+y_3=1$, we have the inequality $$y_1y_2y_3\leq k(x_1y_1+x_2y_2+x_3y_3).$$
My question is what do one have to do in problems of this kind, or in problems with max or min. Should I try find functions ? If anyone has an idea please give me some advice. I posted this problem because I have never seen another one like it. It is supposed to be on the level of 10th grade. It's a problem from Romanian District Olympiad. I was interested in the solution not because I wanted the solution, but I wanted to learn and understand better the idea behind it.
Let $$E={y_1y_2y_3\over x_1y_1+x_2y_2+x_3y_3}$$
we are trying to find a maximum for $E$. This will be achieved when $N= x_1y_1+x_2y_2+x_3y_3$ will be minimal. Because o simmetry we can assume that $y_3\geq y_1,y_2$, so $$ N= (y_1-y_3)x_1+(y_2-y_3)x_2+y_3$$
We see that $N$ is linear on $x_1$ (and on $x_2$). Since $y_1-y_3$ and $y_2-y_3$ are non positive $N$ will be minimal when $x_1$ and $x_2$ will be the most, i.e. $x_1=x_2 = {1\over 2}$ and thus $x_3=0$, so $N_{min} = y_1+y_2+y_3$
Let $t=y_3$. Now we have:
$$E={y_1y_2y_3\over x_1y_1+x_2y_2+x_3y_3} \leq {2y_1y_2t\over 1-t}$$
Since $$ y_1y_2\leq {(y_1+y_2)^2\over 4} = {(1-t)^2\over 4}$$we get $$ E\leq {t(1-t)\over 2}\leq {(t+(1-t))^2\over 8} = {1\over 8}$$
So $k={1\over 8}$.