Prove $x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$ .

Question

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 !

Answer 1

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 $y-x\leq 2\sqrt3$.

Answer 2

Let $y-x=d$

$z=3-x-y=3-x-(x+d)=3-2x-d$

$9=x^2+(x+d)^2+(3-d-2x)^2$

Rearrange to form a quadratic equation in $x$

As $x$ is real, the discriminant must be $\ge0$

Answer 3

Another using calculus:

$(x+y+z)^2 = x^2+y^2+z^2 +2(xy+yz+zx) \implies (xy+yz+zx)=0 \implies xy = -z(x+y)\implies xy=-z(3-z)$

$(y-x)^2 = (x+y)^2 -4xy$

Now $(y-x)^2 = (3-z)^2 +4z(3-z) = 9+6z-3z^2$

Maximum of $(y-x)$ can be found by setting $\frac{d\sqrt{9+6z-3z^2}}{dz} = 0$ so $z = 1$

and Evaluating at $z=1$ $(y-x) \le2\sqrt{3}$

Answer 4

Such an image (this is not an answer) due to a p u