TITLE: If $\{x,y,z\}\subset[-1,1]$ and $x+y+z=0$ so $\sum\limits_{cyc}\sqrt{1+x+\frac{y^2}{6}}\leq3$
QUESTION [18 upvotes]: Let $\{x,y,z\}\subset[-1,1]$ such that $x+y+z=0$. Prove that:
$$\sqrt{1+x+\frac{y^2}{6}}+\sqrt{1+y+\frac{z^2}{6}}+\sqrt{1+z+\frac{x^2}{6}}\leq3$$
I tried C-S, but without success.

REPLY [2 votes]: @cafaxo gave a nice method to eliminate the root signs. Let $\lambda_1 = \frac{2+x}{6}, \ \lambda_2 = \frac{2+y}{6}, \ \lambda_3 = \frac{2+z}{6}.$
It holds that $\lambda_1, \lambda_2, \lambda_3 > 0; \ \lambda_1 + \lambda_2 + \lambda_3 = 1.$ Note that $t\mapsto \sqrt{t}, \ t \ge 0$ is concave.
@cafaxo obtained
\begin{align}
&\sqrt{1 + x + \frac{y^2}{6}} + \sqrt{1 + y + \frac{z^2}{6}} + \sqrt{1 + z + \frac{x^2}{6}}\\
=\ & \lambda_1 \sqrt{\frac{1 + x + \frac{y^2}{6}}{\lambda_1^2}}
+ \lambda_2 \sqrt{\frac{1 + y + \frac{z^2}{6}}{\lambda_2^2}} + \lambda_3 \sqrt{\frac{1 + z + \frac{x^2}{6}}{\lambda_3^2}}\\
\le \ &
\sqrt{\frac{1 + x + \frac{y^2}{6}}{\lambda_1} + \frac{1 + y + \frac{z^2}{6}}{\lambda_2} + \frac{1 + z + \frac{x^2}{6}}{\lambda_3}}\\
= \ & \sqrt{\frac{6 + 6x + y^2}{2+x} + \frac{6 + 6y + z^2}{2+y} + \frac{6 + 6z + x^2}{2+z}}.
\end{align}
It suffices to prove that
$$\frac{6 + 6x + y^2}{2+x} + \frac{6 + 6y + z^2}{2+y} + \frac{6 + 6z + x^2}{2+z} \le 9$$
or (noting that $z = -x-y$)
\begin{align}
&x^4+2 x^3 y+3 x^2 y^2+2 x y^3+y^4-2 x^3+9 x^2 y+15 x y^2+2 y^3\\
&\quad +4 x^2+4 x y+4 y^2 \ge 0 \tag{1}
\end{align}
for $x, y\in [-1, 1]; \ -1\le x + y \le 1.$
My solution
Let me give a different method to prove (1).
With computer, here is a SOS (Sum of Squares) solution:
(1) is true since (note: $A_1, A_2, A_3, A_4$ are all positive semidefinite)
\begin{align}
&x^4+2 x^3 y+3 x^2 y^2+2 x y^3+y^4-2 x^3+9 x^2 y+15 x y^2+2 y^3+4 x^2+4 x y+4 y^2\\
=\ & \frac{1}{60}\Big[u^TA_1u + (1-x)u^TA_2u + (1-y)u^TA_3u + (x+y+1)u^TA_4u\Big]
\end{align}
where
$$u = \left(\begin{array}{c}  x\\  y\\ x^2\\  xy\\ y^2 \end{array}\right), $$
$$
A_1 = \left(\begin{array}{ccccc} 140 & 80 & -30 & 165 & 120\\ 80 & 150 & 55 & 230 & 51\\ -30 & 55 & 60 & 60 & -40\\ 165 & 230 & 60 & 380 & 120\\ 120 & 51 & -40 & 120 & 108 \end{array}\right), \quad A_2 = \left(\begin{array}{ccccc} 80 & 0 & 0 & 0 & 40\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 40 & 0 & 0 & 0 & 20 \end{array}\right),
$$
$$A_3 = \left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 0\\ 0 & 10 & 0 & 0 & 14\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 14 & 0 & 0 & 20 \end{array}\right),
\quad A_4 = \left(\begin{array}{ccccc} 20 & 40 & 0 & 0 & -20\\ 40 & 80 & 0 & 0 & -40\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ -20 & -40 & 0 & 0 & 20 \end{array}\right).$$