How can prove this inequality(8) Let $a,b,c,x,y,z >0$ and $A=a^2+b^2+c^2,\ B=x^2+y^2+z^2,\ C=ax+by+cz$. By Cauchy-Schwarz inequality, we always have $C^2\le AB$. If $C^2<AB$, prove that
$$
\frac{A}{C+\sqrt{2(AB-C^2)}}<\frac{a+b+c}{x+y+z}<\frac{C+\sqrt{2(AB-C^2)}}{B}.
$$
I created this inequality. Are there any nice proofs?

It seems the following.
The inequality is not strict. For instance, put $a=b=c=x=1$, $y=z=0$. Then
$A=3$, $B=C=1$, and $C^2<AB$. But
$$\frac{a+b+c}{x+y+z}=3=\frac{C+\sqrt{2(AB-C^2)}}{B}.$$
The non-strict inequality can be easily proved by means of spherical trigonometry.
Consider the following vectors of unit length in $\mathbb R^3$. Let $u=(a,b,c)/\sqrt{A}$,
$v=(x,y,z)/\sqrt{B}$, and $w=(1,1,1)/\sqrt{3}$. Let $\alpha=\angle (v,u)$, $\beta=\angle(u,w)$,
and $\gamma=\angle(w,v)$. Since $a,b,$ and $c$ are positive, we can easily check that $\cos\beta=(a+b+c)/\sqrt{3A}=(a+b+c)/\sqrt{3(a^2+b^2+c^2)}\ge
1/\sqrt{3}$. Then $\tan\beta\le\sqrt{2}$. By spherical law of cosines, we have that
$\cos\gamma=\cos\alpha\cos\beta+\sin\alpha\sin\beta\cos\angle u$. Then
$\cos\gamma/\cos\beta\le \cos\alpha+\sin\alpha\tan\beta\le\cos\alpha+\sqrt{2}\sin\alpha$.
Substituting the cosines by the inner products, we obtain
$$\frac {(x+y+z)/\sqrt{3B}}{(a+b+c)/\sqrt{3A}}\le\frac{C}{\sqrt{AB}}+\sqrt{2-2\left(\frac{C}{\sqrt{AB}}\right)^2}.$$
The equality should hold only if $\cos\angle u=1$  (that is, the vectors $u,v,w$ and $0$ are
coplanar) and $\cos\beta=1/\sqrt{3}$, that is one of $a,b,$ and $c$ is equal to $1$, and the others
are equal to $0$.
From the another law of cosines
$\cos\beta=\cos\alpha\cos\gamma+\sin\alpha\sin\gamma\cos\angle v$,
we can similarly obtain the inequality
$$\frac{a+b+c}{x+y+z}\le \frac{C+\sqrt{2(AB-C^2)}}{B}.$$
PS. Maybe the $n$-dimensional version of the inequality holds, if we change in it "$2$" to "$n-1$".
