Open Contests

$$\begin{array}{rl}252& =3x+7y+14z\ge 3\sqrt[3]{3x\cdot 7y\cdot 14z}=3\sqrt[3]{3\cdot 7\cdot 14\cdot (2016+u^2)}\\ & \ge 3\sqrt[3]{3\cdot 7\cdot 14\cdot 2016}=3\sqrt[3]{2^6\cdot 3^3\cdot 7^3}=2^2\cdot 3^2\cdot 7=252.\end{array}$$ To not get a contradiction, we must have equality in both inequalities, hence $3x=7y=14z$ and $u=0$. From the first equation of the system we finally obtain $3x=7y=14z=\frac{252}{3}=84$, hence $x=28,y=12$ and $z=6$.

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Alternative solution.

The second equation implies $xyz\ge 2016$. Substituting the value of $x$ from the first equation here gives $(252-7y-14z)yz\ge 6048$, which is equivalent to $(36-y-2z)yz\ge 864$. We find the maximum of function $f(y,z)=(36-y-2z)yz$. Fixing $z>0$ arbitrarily, we obtain $\frac{df(y,z)}{dy}=36z-2yz-2z^2=2z(18-y-z)$. The partial derivative with respect to $y$ is zero, if $z=18-y$. As the second derivative is negative, the function $f(y,z)$ has exactly one maximum at $y=18-z$ for any fixed positive number $z$. Define $g(z)=f(18-z,z)=(18-z{)}^{2}z$. Its derivative $g^{\prime }(z)=(18-z)(18-3z)$ is zero at $z=6$ ($z=18$ does not count since then $y=0$). As the second derivative is negative at $z=6$, the extremum found is a maximum again. This means that out of all partial maxima of $f(y,z)$, the one for $z=6$ is the largest. As $g(6)=f(12,6)=864$, the initial system of equations can be satisfied only if $y=12$ and $z=6$. Substituting these values into the initial system leads to $3x=84,72x-u^2=2016$. The only solution of this system is $(x,u)=(28,0)$.