jmc

Suppose equality occurs when $(x,y,z)=(x_0,y_0,z_0).$ To find and prove the minimum value, it looks like we're going to have to put together some inequalities like $$x^2+y^2\ge 2xy.$$Remembering that equality occurs when $x=x_0$ and $y=y_0,$ or $\frac{x}{x_0}=\frac{y}{y_0}=1,$ we form the inequality $$\frac{x^2}{x_0^2}+\frac{y^2}{y_0^2}\ge \frac{2xy}{x_0{y}_0}.$$Then $$\frac{y_0}{2x_0}\cdot x^2+\frac{x_0}{2y_0}\cdot y^2\ge xy.$$Similarly, $$\begin{array}{r}\frac{z_0}{2x_0}\cdot x^2+\frac{x_0}{2z_0}\cdot z^2\ge xz,\\ \frac{z_0}{2y_0}\cdot y^2+\frac{y_0}{2z_0}\cdot z^2\ge xz.\end{array}$$Adding these, we get $$\frac{y_0+z_0}{2x_0}\cdot x^2+\frac{x_0+z_0}{2y_0}\cdot y^2+\frac{x_0+y_0}{2z_0}\cdot z^2\ge xy+xz+yz.$$We want to maximize $10x^2+10y^2+z^2,$ so we want $x_0,$ $y_0,$ and $z_0$ to satisfy $$\frac{y_0+z_0}{x_0}:\frac{x_0+z_0}{y_0}:\frac{x_0+y_0}{z_0}=10:10:1.$$Let $$\begin{array}{rl}{y}_0+z_0& =10k{x}_0,\\ x_0+z_0& =10k{y}_0,\\ x_0+y_0& =k{z}_0.\end{array}$$Then $$\begin{array}{rl}{x}_0+y_0+z_0& =(10k+1)x_0,\\ x_0+y_0+z_0& =(10k+1)y_0,\\ x_0+y_0+z_0& =(k+1)z_0.\end{array}$$Let $t=x_0+y_0+z_0.$ Then $x_0=\frac{t}{10k+1},$ $y_0=\frac{t}{10k+1},$ and $z_0=\frac{t}{k+1},$ so $$\frac{t}{10k+1}+\frac{t}{10k+1}+\frac{t}{k+1}=t.$$Hence, $$\frac{1}{10k+1}+\frac{1}{10k+1}+\frac{1}{k+1}=1.$$This simplifies to $10k^2-k-2=0,$ which factors as $(2k-1)(5k+2)=0.$ Since $k$ is positive, $k=\frac{1}{2}.$

Then $x_0=\frac{t}{6},$ $y_0=\frac{t}{6},$ and $z_0=\frac{2t}{3}.$ Substituting into $xy+xz+yz=1,$ we get $$\frac{t^2}{36}+\frac{t^2}{9}+\frac{t^2}{9}=1.$$Solving, we find $t=2,$ and the minimum value of $10x^2+10y^2+z^2$ is $$10\cdot \frac{t^2}{36}+10\cdot \frac{t^2}{36}+\frac{4t^2}{9}=t^2=\boxed{4}.$$