If $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}?$ True or false?
If $x \geq 0, y \geq 0, z \geq 0 $ and $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}.$
I want to know if there is a way to demonstrate this conditional inequality.
I know I can make a connection with two properties known in a triangle. I tried to find an algebraic demonstration of these problems and we did.
Thanks in advance for any suggestions.

Let $(a,b,c)=(\lambda x,\lambda y,\lambda z)$, where $\lambda(x+y+z)=\frac32$. Then $a+b+c=\frac32$, and we have
$$a^2+b^2+c^2+2abc=\lambda^2(x^2+y^2+z^2+2\lambda xyz),$$
which is increasing in $\lambda$. Hence to show $\lambda\geq1$ it suffices to prove
$$a+b+c=\frac32\implies a^2+b^2+c^2+2abc\geq1.$$
We now use the notation for cyclic sums $\sum$ and symmetric sums $\sum_\text{sym}$. Note that $(\sum a)^2=\sum a^2+2\sum ab=\frac94$, so
\begin{align*}
\sum a^2+2\sum abc&\geq1\\
\iff\frac94\left(\sum a^2+2\sum abc\right)&\geq\sum a^2+2\sum ab\\
\iff18abc&\geq8\sum ab-5\sum a^2\\
\iff27abc&\geq\left(\sum a\right)\left(8\sum ab-5\sum a^2\right)\\
&=8\sum_\text{sym}a^2b+24abc-5\sum a^3-5\sum_\text{sym}a^2b\\
\iff5\sum a^3+3abc&\geq3\sum_\text{sym}a^2b.
\end{align*}
The last inequality follows from AM-GM ($\sum a^3\geq3abc$) and Schur's inequality ($\sum a^3+3abc\geq3\sum_\text{sym}a^2b$):
$$5\sum a^3+3abc\geq3\left(\sum a^3+3abc\right)\geq3\sum_\text{sym}a^2b,$$
and we are done.
