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

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.

Answer 1

Let $Z = \{\,(x,y,z) \in [-1,1]^3 \mid x+y+z=0\,\}$.

Let $f \colon Z \rightarrow \mathbb{R}$ be the function given by $$f(x,y,z) = \sqrt{1+x+\frac{y^2}{6}} + \sqrt{1+y+\frac{z^2}{6}} + \sqrt{1+z+\frac{x^2}{6}}.$$

Let $g \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function given by \begin{align*} g(x,y) =\ &x^4+2 x^3 y-2 x^3+3 x^2 y^2+9 x^2 y+4 x^2 \\ &+2 x y^3+15 x y^2+4 x y+y^4+2 y^3+4 y^2. \end{align*}

Lemma 1. If $(x,y,z) \in Z$ and $g(x,y) \geq 0$, then $f(x,y,z) \leq 3$.

Proof. Let $(x,y,z) \in Z$ such that $g(x,y) \geq 0$. Define $\lambda_1 = \frac{x+2}{6}$, $\lambda_2 = \frac{y+2}{6}$ and $\lambda_3 = \frac{z+2}{6}$. We have $\lambda_1,\lambda_2,\lambda_3 \geq 0$ and $\lambda_1+\lambda_2+\lambda_3=1$. Since $\alpha \mapsto \sqrt{\alpha}$ is concave, we have by Jensen's inequality: \begin{align*} f(x,y,z) &= \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{6 x+y^2+6}{6 \lambda_1 ^2}} + \lambda_2 \sqrt{\frac{6 y+z^2+6}{6 \lambda_2 ^2}} + \lambda_3 \sqrt{\frac{6 z+x^2+6}{6 \lambda_3 ^2}} \\ &\leq \sqrt{ \lambda_1 \frac{6 x+y^2+6}{6 \lambda_1 ^2} + \lambda_2 \frac{6 y+z^2+6}{6 \lambda_2 ^2} + \lambda_3 \frac{6 z+x^2+6}{6 \lambda_3 ^2}} \\ &= \sqrt{ \frac{6 x+y^2+6}{x+2} + \frac{6 y+z^2+6}{y+2} + \frac{6 z+x^2+6}{z+2}}. \end{align*} We rewrite the fractions inside the square root to have the same denominator and then replace $z$ with $-x-y$: \begin{align*} &(9 - f(x,y,z)^2) (x+2) (y+2) (z+2) \\&\geq \left(9-\frac{6 x+y^2+6}{x+2} - \frac{6 y+z^2+6}{y+2} - \frac{6 z+x^2+6}{z+2}\right) (x+2) (y+2) (z+2) \\&= g(x,y) \geq 0. \end{align*} $$\tag*{$\Box$}$$

Let $\widetilde{Z} = \{\,(x,y,z) \in Z \mid z \leq x \land z \leq y\,\}$.

Lemma 2. If $(x,y,z) \in \widetilde{Z}$, then $-3 x^2+6 x+4 y \geq 0$.

Proof. Let $(x,y,z) \in \widetilde{Z}$. If $x \geq 0$, then we have: \begin{align*} -3 x^2+6 x+4 y &= 3x(1+(1-x))+4 y \\&\geq 3x+4y \\&\geq 2x+4y \\&\geq 2x+2y+2z = 0. \end{align*} If $x < 0$, then we have: \begin{align*} -3 x^2+6 x+4 y &= -3x^2 -2x + 4y + 8x \\&\geq -3x^2 -2x + 4y + 4x + 4z \\&= -x(3x+2) \\&\geq -2x(2x+1) \\&\geq -2x(x+z+1) \\&= -2x(-y+1) \geq 0. \end{align*} $$\tag*{$\Box$}$$

Let $\widetilde{g} \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function given by \begin{align*} \widetilde{g}(x,y) &= g(x,y)-y^2(-3 x^2+6 x+4 y)-2 (1-x) (1-y) (x+y)^2 \\ &= x^4 + 2 x^2 y^2 + y^4 + 15 x^2 y + 15 x y^2 + 2 x^2 + 2 y^2. \end{align*}

Corollary 2.1. If $(x,y,z) \in \widetilde{Z}$ and $\widetilde{g}(x,y) \geq 0$, then $g(x,y) \geq 0$ and by Lemma 1 we have $f(x,y,z) \leq 3$.

Let $\hat{Z} = \{\,(x,y,z) \in \widetilde{Z} \mid x > 0\,\}$.

Let $\hat{g} \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function given by \begin{align*} \hat{g}(x,y) &= \frac{1}{x}\left(\widetilde{g}(x,y)-(1 - x) (2 y^2 + y^4) - x y^4\right) \\ &= x^3+2 x y^2+15 x y+2 x+17 y^2 \\ &= (17+2x) y^2 + 15 x y + x^3 + 2 x. \end{align*}

Lemma 3. If $\hat{g}(x,y) \geq 0$ for all $(x,y,z) \in \hat{Z}$, then $f(x,y,z) \leq 3$ for all $(x,y,z) \in Z$.

Proof. Let $(x,y,z) \in Z$.

Then there exists $(\widetilde{x},\widetilde{y},\widetilde{z}) \in \{\,(x,y,z),(y,z,x),(z,x,y)\,\} \cap \widetilde{Z}$.

If $\widetilde{x} = \widetilde{y} = \widetilde{z} = 0$, then $f(x,y,z) = f(0,0,0) = 3$.

If $\widetilde{z} < 0$, then there exists $(\hat{x},\hat{y},\hat{z}) \in \{\,(\widetilde{x},\widetilde{y},\widetilde{z}), (\widetilde{y},\widetilde{x},\widetilde{z})\,\} \cap \hat{Z}$.

Since $\hat{g}(\hat{x},\hat{y}) \geq 0$ and $\hat{x} > 0$, we have $\widetilde{g}(\hat{x},\hat{y}) \geq 0$ by the definition of $\hat{g}$. Hence $\widetilde{g}(\widetilde{x},\widetilde{y}) = \widetilde{g}(\hat{x},\hat{y}) \geq 0$. By Corollary 2.1 we have $ f(\widetilde{x},\widetilde{y},\widetilde{z}) \leq 3$, hence $f(x,y,z) = f(\widetilde{x},\widetilde{y},\widetilde{z}) \leq 3$. $$\tag*{$\Box$}$$

Lemma 4. If $(x,y,z) \in \hat{Z}$, then $\hat{g}(x,y) > 0$.

Proof. Let $(x,y,z) \in \hat{Z}$. The function $\mu \mapsto \hat{g}(x,\mu)$ is quadratic with $17+2x > 0, 15x > 0$, therefore it has a minimum point at $\mu = -\frac{15x}{34+4x}$.

Let $h \colon [0,1] \rightarrow \mathbb{R}$ be the function given by \begin{align*} h(\lambda) &= \frac{4 (17 + 2 \lambda)}{\lambda} \cdot \hat{g}(\lambda, -\frac{15\lambda}{34+4\lambda}) \\&= 8 \lambda^3+68 \lambda^2-209 \lambda+136. \end{align*} We have $h'(\lambda) = 24 \lambda^2+136 \lambda-209 < 0$ for all $\lambda \in [0,1]$. Therefore, $h$ is monotonically decreasing. Since $h(1) = 3$, $h$ is positive.

We have: \begin{align*} \hat{g}(x,y) &\geq \hat{g}(x, -\frac{15x}{34+4x}) \\&= \frac{x}{4 (17 + 2 x)} \cdot h(x) \\&> 0. \end{align*} $$\tag*{$\Box$}$$

Corollary 4.1. By Lemma 3 and Lemma 4, we have that $f(x,y,z) \leq 3$ for all $(x,y,z) \in Z$.

Answer 2

@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 $$u = \left(\begin{array}{c} x\\ y\\ x^2\\ xy\\ y^2 \end{array}\right), $$ $$ A_1 = \left(\begin{array}{ccccc} 140 & 80 & -30 & 165 & 120\\ 80 & 150 & 55 & 230 & 51\\ -30 & 55 & 60 & 60 & -40\\ 165 & 230 & 60 & 380 & 120\\ 120 & 51 & -40 & 120 & 108 \end{array}\right), \quad A_2 = \left(\begin{array}{ccccc} 80 & 0 & 0 & 0 & 40\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 40 & 0 & 0 & 0 & 20 \end{array}\right), $$ $$A_3 = \left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 0\\ 0 & 10 & 0 & 0 & 14\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 14 & 0 & 0 & 20 \end{array}\right), \quad A_4 = \left(\begin{array}{ccccc} 20 & 40 & 0 & 0 & -20\\ 40 & 80 & 0 & 0 & -40\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ -20 & -40 & 0 & 0 & 20 \end{array}\right).$$