Prove $x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$ . Prove
$$x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$$
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
We have
$$(\,x+ y+ z\,)^{\,2}+ (\,-\,x+ y+ z\,)^{\,2}+ (\,x- y+ z\,)^{\,2}+ (\,x+ y- z\,)^{\,2}= 4(\,x^{\,2}+ y^{\,2}+ z^{\,2}\,)= 36$$
Or
$$\left ( z+ (\,y- x\,) \right )^{\,2}+ \left ( z- (\,y- x\,) \right )^{\,2}+ (\,3- 2\,z\,)^{\,2}= 27$$
Or
$$3\,z^{\,2}- 6\,z+ (\,y- x\,)^{\,2}= 9$$
Or
$$(\,y- x\,)^{\,2}= -\,3(\,z- 1\,)^{\,2}+ 12\leqq 12\,\therefore\,y- x\leqq |\,y- x\,|\leqq 2\sqrt{3}$$
Q.E.D. The equality condition occurs when $z= 1\,\therefore\,x+ y= 2\,\therefore\,x= 1- \sqrt{3},\,y= 1+ \sqrt{3}$.
Say it (Added)
The @user10354138's solution is so amazing, I try writing the inequality into the homonogeous form, then find $t\!=\!constant$ such that $3(\!y- z\!)^{\!2}\leqq 2t(\!x+ y+ z\!)^{\!2}+ 2(\!1- t\!)(\!x^{\!2}+ y^{\!2}+ z^{\!2}\!)$. That will lead to:
$${\rm discriminant}= 0\,\therefore\,t= -\,2,\,-\,\frac{1}{2},\,1$$
The coefficients of $y^{2}$ and $z^{2}$ both are negative there, I can't make the form like the solution as follow !

Since $\frac1{\sqrt3}(1,1,1), \frac1{\sqrt2}(-1,1,0), \frac1{\sqrt6}(1,1,-2)$ forms an orthonormal basis of $\mathbb{R}^3$ with respect to the usual inner product, we have
$$
x^2+y^2+z^2=\frac13(x+y+z)^2+\frac12(y-x)^2+\frac16(x+y-2z)^2
$$
So
$$
\frac12(y-x)^2\leq (x^2+y^2+z^2)-\frac13(x+y+z)^2=9-3=6
$$
and hence $y-x\leq 2\sqrt3$.
