Prove $\sum\limits_{\mathrm{cyc}} \sqrt{a^3+2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)}$ for $a, b, c \ge 0$

Question

Let $a,b,c\ge 0$,show that $$\sqrt{a^3+2}+\sqrt{b^3+2}+\sqrt{c^3+2}\ge \sqrt{\dfrac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)}$$

Answer 1

Some thoughts:

WLOG, assume that $a \le b \le c$.

By AM-GM, we have $$\sqrt{\dfrac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)} \le \frac{(9 + 3\sqrt 3)(a^2 + b^2 + c^2)}{2\sqrt{3}\, c + 6 + 10\sqrt 3} + \frac{2\sqrt{3}\, c + 6 + 10\sqrt 3}{8}.$$

It suffices to prove that $$\sqrt{a^3 + 2} + \sqrt{b^3 + 2} + \sqrt{c^3 + 2} \ge \frac{(9 + 3\sqrt 3)(a^2 + b^2 + c^2)}{2\sqrt{3}\, c + 6 + 10\sqrt 3} + \frac{2\sqrt{3}\, c + 6 + 10\sqrt 3}{8}$$ or \begin{align*} &\left(\sqrt{a^3 + 2} - \frac{(9 + 3\sqrt 3)a^2}{2\sqrt{3}\, c + 6 + 10\sqrt 3} \right) + \left(\sqrt{b^3 + 2} - \frac{(9 + 3\sqrt 3) b^2 }{2\sqrt{3}\, c + 6 + 10\sqrt 3}\right)\\[6pt] &\qquad + \sqrt{c^3 + 2} - \frac{(9 + 3\sqrt 3)c^2}{2\sqrt{3}\, c + 6 + 10\sqrt 3} - \frac{2\sqrt{3}\, c + 6 + 10\sqrt 3}{8} \ge 0. \tag{1} \end{align*} Observing the expressions inside the first and the second parentheses, clearly, we only need to prove the case that $a = b$. Thus, it suffices to prove that, for all $0 \le a \le c$, \begin{align*} F(a, c)&:= 2 \cdot \left(\sqrt{a^3 + 2} - \frac{(9 + 3\sqrt 3)a^2}{2\sqrt{3}\, c + 6 + 10\sqrt 3} \right)\\[6pt] &\qquad + \sqrt{c^3 + 2} - \frac{(9 + 3\sqrt 3)c^2}{2\sqrt{3}\, c + 6 + 10\sqrt 3} - \frac{2\sqrt{3}\, c + 6 + 10\sqrt 3}{8} \ge 0. \tag{2} \end{align*} This inequality is true which is verified by Mathematica.

By the way, we can find the minimum of $$f(a) := \sqrt{a^3 + 2} - \frac{(9 + 3\sqrt 3)a^2}{2\sqrt{3}\, c + 6 + 10\sqrt 3}$$ in closed form (radical form or trigonometric form). However it is quite complicated.

Answer 2

Remarks: Here is an alternative proof using derivatives.

First of all, we give the following auxiliary result. The proof is given at the end.

Fact 1: Let $q\ge 0$ be fixed. If $(x_0, y_0)$ is a minimizer of $\sqrt{x^3 + 2} + \sqrt{y^3 + 2}$ subject to $x, y\ge 0$ with $x^2 + y^2 = q$, then $(x_0 - y_0)(x_0^2y_0^2 - 2x_0 - 2y_0) = 0$.

Now, consider the minimum of $$f(a, b, c) := \sqrt{a^3+2}+\sqrt{b^3+2}+\sqrt{c^3+2} - \sqrt{\frac{9+3\sqrt{3}}{2}(a^2+b^2+c^2)},$$ subject to $a, b, c \ge 0$. We have $f(1, 1, 1 + \sqrt 3) = 0$.

Let $(a_0, b_0, c_0)$ be a minimizer. We have $(a_0 - b_0)^2 + (b_0 - c_0)^2 + (c_0 - a_0)^2 \ne 0$ since $f(t, t, t) \ge 3\sqrt{t^3 + 2} - \sqrt{24t^2} > 0$.

By Fact 1, we have \begin{align*} (a_0 - b_0)(a_0^2b_0^2 - 2a_0 - 2b_0) &= 0, \tag{1}\\ (b_0 - c_0)(b_0^2c_0^2 - 2b_0 - 2c_0) &= 0, \tag{2}\\ (c_0 - a_0)(c_0^2a_0^2 - 2c_0 - 2a_0) &= 0. \tag{3} \end{align*} (Note: For example, letting $c = c_0$ and $a^2 + b^2 = a_0^2 + b_0^2$, we know that $(a_0, b_0)$ is the minimizer of $\sqrt{a^3+2} + \sqrt{b^3 + 2}$ subject to $a, b\ge 0$ with $a^2+b^2 = a_0^2 + b_0^2$. By Fact 1, we have (1).)

From (1)-(3), we have $a_0 = b_0, c_0 = (1 + \sqrt{2b_0^3 + 1})/b_0^2$ (or permutations). The proof is easy and thus omitted.

We have \begin{align*} &f\left(b, b, \frac{1 + \sqrt{2b^3 + 1}}{b^2}\right) \\[6pt] ={}& \frac{1 + \sqrt{2b^3 + 1}}{2b^3}\left(2\sqrt{(b^3+2)(2b^3 + 1)} - b\sqrt{(9+3\sqrt 3)\left(2b^3 + 4 - 2\sqrt{2b^3+1}\right)}\right)\\[6pt] \ge{}& 0. \end{align*} (Note: It suffices to prove that $2^6(b^3+2)^3(2b^3 + 1)^3 - b^6(9+3\sqrt 3)^3\left(2b^3 + 4 - 2\sqrt{2b^3+1}\right)^3\ge 0$. Then use $u = \sqrt{2b^3 + 1}-1$ to eliminate the square root sign. The rest is smooth.)

Thus, we have $f(a, b, c) \ge 0$ on $a, b, c \ge 0$.

We are done.

$\phantom{2}$

---

Proof of Fact 1:

The non-trivial case is $q > 0$.

Consider the function $$f(x) := \sqrt{x^3+2} + \sqrt{(q - x)^{3/2} + 2}, \quad x\in [0, \sqrt{q}].$$

We have $$f'(x) = \frac{3x^2}{2\sqrt{x^3+2}} - \frac{3x\sqrt{q - x^2}}{2\sqrt{(q-x^2)^{3/2} + 2}}.$$

We have $f'(\sqrt{q}) = \frac{3q}{2\sqrt{q^{3/2} + 2}} > 0$. Also, we have $f(0) = f(\sqrt{q})$. Thus, the minimum of $f(x)$ on $[0, \sqrt{q}]$ occurs at the stationary point(s) in the interior of the interval. From $f'(x) = 0$ and $0 < x < \sqrt{q}$, we have $$x^2(q - x^2)^{3/2} + x^5 + 4x^2 - qx^3 - 2q = 0$$ which is written as (note: $y = \sqrt{q - x^2}$) $$(x - y)(x^2 y^2 - 2x - 2y) = 0.$$

We are done.

Answer 3

Sketch of a proof.

Letting $x = a^2, y = b^2, z = c^2$, it suffices to prove that, for all $x, y, z \ge 0$, $$\sqrt{x^{3/2}+2}+\sqrt{y^{3/2}+2}+\sqrt{z^{3/2}+2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(x+y+z)}. \tag{1}$$

Let $f(u) := \sqrt{u^{3/2} + 2}$. Then, $f(u)$ is convex on $[0, 2^{4/3}]$, and concave on $[2^{4/3}, \infty)$ (easy to prove).

WLOG, assume that $x \le y \le z$.

We split into three cases.

Case 1: $y \le 2^{4/3}$

By Jensen's inequality, it suffices to prove that $$2\sqrt{\left(\frac{x + y}{2}\right)^{3/2} + 2} + \sqrt{z^{3/2} + 2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(x+y+z)}. \tag{2}$$

Letting $u = \sqrt{\frac{x + y}{2}}$ and $z = v^2$, it suffices to prove that, for all $u, v \ge 0$,
$$2\sqrt{u^3 + 2} + \sqrt{v^3 + 2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(2u^2 + v^2)}. \tag{3}$$

(3) is true. We can prove it by Calculus. Specifically, we can find all $u, v > 0$ such that $\frac{\partial F}{\partial u} = \frac{\partial F}{\partial v} = 0 $. Actually, there is a proof without calculus.

By the way, Michael Rozenberg's idea works. By Holder, it suffices to prove that, for all $u, v \ge 0$, \begin{align*} \frac{(2u^2 + v^2 + 3 + 3\sqrt 3)^3} {2\cdot \frac{(u^2 + 1 + \sqrt 3)^3}{u^3 + 2} + \frac{(v^2 + 1 + \sqrt 3)^3}{v^3 + 2}} \ge \frac{9 + 3\sqrt 3}{2}(2u^2 + v^2). \end{align*} This inequality is true which is verified by Mathematica. But I don't have a nice proof.

Case 2: $x \le 2^{4/3} < y$

By Karamata's inequality, since $(y, z)$ is majorized by $(y + z - 2^{4/3}, 2^{4/3})$, we have $$\sqrt{y^{3/2} + 2} + \sqrt{z^{3/2} + 2} \ge \sqrt{(y + z - 2^{4/3})^{3/2} + 2} + \sqrt{(2^{4/3})^{3/2} + 2}.$$

By Jensen's inequality, we have $$\sqrt{x^{3/2} + 2} + \sqrt{(2^{4/3})^{3/2} + 2} \ge 2\sqrt{\left(\frac{x + 2^{4/3}}{2}\right)^{3/2} + 2}.$$

It suffices to prove that $$2\sqrt{\left(\frac{x + 2^{4/3}}{2}\right)^{3/2} + 2} + \sqrt{(y + z - 2^{4/3})^{3/2} + 2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(x+y+z)}. \tag{4}$$

Letting $u = \sqrt{\frac{x + 2^{4/3}}{2}}$ and $v = \sqrt{y + z - 2^{4/3}}$, it suffices to prove that, for all $u, v \ge 0$, $$2\sqrt{u^3 + 2} + \sqrt{v^3 + 2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(2u^2 + v^2)}. \tag{5}$$ (5) is the same as (3) in Case 1.

Case 3: $2^{4/3} < x$

By Karamata's inequality, since $(x, y, z)$ is majorized by $(2^{4/3}, 2^{4/3}, x + y + z - 2\cdot 2^{4/3})$, we have \begin{align*} &\sqrt{x^{3/2} + 2} + \sqrt{y^{3/2} + 2} + \sqrt{z^{3/2} + 2}\\ \ge{}& 2\sqrt{(2^{4/3})^{3/2} + 2} + \sqrt{(x + y + z - 2\cdot 2^{4/3})^{3/2} + 2}. \end{align*}

It suffices to prove that $$2\sqrt{(2^{4/3})^{3/2} + 2} + \sqrt{(x + y + z - 2\cdot 2^{4/3})^{3/2} + 2} \ge \sqrt{\frac{9+3\sqrt{3}}{2}(x + y + z)}. \tag{6}$$

Letting $u = \sqrt{x + y + z - 2\cdot 2^{4/3}}$, it suffices to prove that, for all $u \ge 2^{2/3}$, $$F(u) := 2\sqrt{6} + \sqrt{u^3 + 2} - \sqrt{\frac{9+3\sqrt{3}}{2}(u^2 + 2\cdot 2^{4/3})} \ge 0.\tag{7}$$ It is not difficult to prove that $F(u) > 0$ for all $u \ge 2^{2/3}$. Also, $F(2^{2/3}) > 0$. Thus, $F(u) > 0$ for all $u \ge 2^{2/3}$.

Answer 4

I think the following will help.

By Holder $$\left(\sum_{cyc}\sqrt{a^3+2}\right)^2\sum_{cyc}\frac{(a^2+1+\sqrt3)^3}{a^3+2}\geq(a^2+b^2+c^2+3+3\sqrt3)^3.$$ Thus, it's enough to prove that $$(a^2+b^2+c^2+3+3\sqrt3)^3\geq\frac{9+3\sqrt3}{2}(a^2+b^2+c^2)\sum_{cyc}\frac{(a^2+1+\sqrt3)^3}{a^3+2},$$ which looks true, but I have no the proof.

At least, we deleted radicals of variables and the inequality is a polynomial inequality already!

Answer 5

Let us consider the task of minimizing of function $$f(a,b,c) = \sum\limits_\bigcirc\sqrt{a^3+2} - \sqrt{\frac{9+3\sqrt3}2}\sqrt{a^2+b^2+c^2}$$ for all the positive $a,b,c.$

The least value of $f(a,b,c)$ can be achieved on the bounds of area, or in its stationary points.

$\color{brown}{\textbf{The bounds.}}$

Easy to see that $f(x,y,z)\to + \infty >0$ if $\underline{(a\to \infty) \vee (b\to \infty) \vee (c\to \infty)}.$

If $\underline{a=b=0}$ then $$f(0,0,c) = 2\sqrt2+\sqrt{c^3+2}-\sqrt{\frac{9+3\sqrt3}2}c\ge 4+\sqrt{c^3+2}-\frac83c.$$ Taking in account that $f(0,0,3) = 2\sqrt2+\sqrt{29}-8>\frac15$ and $$f'(0,0,c) \ge\frac32 \frac{c^2}{\sqrt{c^3+2}}-8,\quad f'(0,0,3)\in\left(-\frac16, -\frac16+1/250\right),\quad f'(0,0,4)\ge0,$$ easy to prove that $$f'(0,0,c)\ge p(c)\ge0,$$ where $$p(c) = \begin{cases} \frac15-\frac16(c-3),\quad\text{if}\quad c\in\left(0,\frac{21}5\right)\\ 0,\quad\text{if}\quad c\in\left(\frac{21}5, \infty\right) \end{cases}$$ (see also Wolfram Alpha plot)

If $\underline{a=0}$ then $$g(b,c) = f(0,b,c) = \sqrt2+\sqrt{b^3+2} + \sqrt{c^3+2} - \sqrt{\frac{9+3\sqrt3}2}\sqrt{b^2+c^2}.$$ The system for the stationary points on this bound is $g'_b = g'_c=0,$ or \begin{cases} \dfrac32\dfrac{b^2}{\sqrt{b^3+2}}-\sqrt{\dfrac{9+3\sqrt3}2}\dfrac b{\sqrt{b^2+c^2}}=0\\ \dfrac32\dfrac{c^2}{\sqrt{c^3+2}}-\sqrt{\dfrac{9+3\sqrt3}2}\dfrac c{\sqrt{b^2+c^2}}=0, \end{cases} \begin{cases} 3b^2(b^2+c^2)=2(3+\sqrt3)(b^3+2)\\ 3c^2(b^2+c^2)=2(3+\sqrt3)(c^3+2), \end{cases} \begin{cases} (c^3+2)b^2-(b^3+2)c^2 = 0\\ 3(b^2+c^2)^2-2(3+\sqrt3)(b^3+c^3+4)=0, \end{cases} \begin{cases} (c-b)(b^2c^2-2(b+c)) = 0\\ 3(b^2+c^2)^2-2(3+\sqrt3)(b^3+c^3+4)=0, \end{cases} $$\left[\begin{align} &\begin{cases} b=c\\ 12c^4-4(3+\sqrt3)(c^3+2)=0 \end{cases}\\ &\begin{cases} b^2c^2=2(b+c),\\ 3(b^2+c^2)^2-2(3+\sqrt3)(b^3+c^3+4)=0, \end{cases}\\ \end{align}\right.\tag1$$ The first system of $(1)$ leads to the solution $$\begin{pmatrix}a\\b\\c\\f\end{pmatrix}= \begin{pmatrix}0\\1.982328\\1.982328\\0.202974\end{pmatrix}. \tag2$$ with the positive values of $f(a,b,c).$

Substitutions $$s=b+c,\quad p=bc$$ change the second system of $(1)$ to the form of \begin{cases} 3(s^2-2p)^2=2(3+\sqrt3)(s^3-3ps+4)\\ p^2=2s, \end{cases} \begin{cases} 3(p^4-8p)^2 - 4(3+\sqrt3)(p^6-12p^3+32) = 0\\ 2s= p^2, \end{cases}

\begin{cases} (p^3-8)(3p^5-24p^2-4(3+\sqrt3)(p^3-4))=0\\ s=\dfrac{p^2}2\\ \{b,c\}=\dfrac{p^2\pm\sqrt{p^4-16p}}4, \end{cases} with the real solutions $$ p \in \{1.340309, 2, 2.832528\},$$

$$\begin{pmatrix}a\\b\\c\\f\end{pmatrix}= \begin{pmatrix} 0 \\ 3.097005 \\ 0.914602 \\ 0.104369 \end{pmatrix} \begin{pmatrix} 0 \\ 0.914602 \\ 3.097005 \\ 0.104369 \end{pmatrix}.\tag3$$

Therefore, the values of $g(b,c)$ on the stationary points with the positive coordinates are positive, i.e. the inequality is satisfied on the bounds.

$\color{brown}{\textbf{The inner stationary points.}}$

The inner stationary points of the function $$f(a,b,c) = \sum\limits_\bigcirc\sqrt{a^3+2} - \sqrt{\frac{9+3\sqrt3}2}\sqrt{a^2+b^2+c^2}$$ can be obtained from the system $f'_a=f'_b=f'_c=0,$ or \begin{cases} \dfrac32\dfrac{a^2}{\sqrt{a^3+2}} = \sqrt{\dfrac{9+3\sqrt3}2}\dfrac a{\sqrt{a^2+b^2+c^2}}\\ \dfrac32\dfrac{b^2}{\sqrt{a^3+2}} = \sqrt{\dfrac{9+3\sqrt3}2}\dfrac b{\sqrt{a^2+b^2+c^2}}\\ \dfrac32\dfrac{c^2}{\sqrt{a^3+2}} = \sqrt{\dfrac{9+3\sqrt3}2}\dfrac c{\sqrt{a^2+b^2+c^2}}, \end{cases} $$\dfrac{a^2}{a^3+2} = \dfrac{b^2}{b^3+2} = \dfrac{c^2}{c^3+2} = \frac23\dfrac{3+\sqrt3}{a^2+b^2+c^2}.$$ Easy to show that the function $h(t)=\frac{t^2}{t^3+2}$ has the maximum in the point $t_m=\sqrt[3]4$ and has the increasing branch on the interval $(0,t_m)$ and decreasing branch on the interval $(t_m,\infty)$ (see also Wolfram Alpha plot). In the accordance with the Dirichlet principle, let WLOG $a=b,$

then $$\dfrac{b^2}{b^3+2} = \dfrac{c^2}{c^3+2} = \frac23\dfrac{3+\sqrt3}{2b^2+c^2},$$ \begin{cases} b^2(c^3+2)-c^2(b^3+2)=0\\ c^2=2(3+\sqrt3)\dfrac{b^3+2}{3b^2}-2b^2, \end{cases} $$\left[\begin{align} &\begin{cases} b=c\\ 9c^4-2(3+\sqrt3)(c^3+2)=0 \end{cases}\\ &\begin{cases} b^2c^2=2(b+c)\\ c=\dfrac{3+\sqrt3}3(b^3+2)-b^4-b \end{cases} \end{align}\right.$$ The first system has the solution $$a=b=c \approx1.582375,\quad f(a,b,c)\approx0.0232638>0.$$ The substitution $$q=c^2$$ allows to present the second system in the form of \begin{cases} \left(\frac12b^2q-2b\right)^2=q\\ q=2(3+\sqrt3)\dfrac{b^3+2}{3b^2}-2b^2\\ c=\dfrac{3+\sqrt3}3(b^3+2)-b(b^3+1), \end{cases} and this leads to the positive solutions $$\begin{pmatrix}a\\b\\c\\f\end{pmatrix}=\left\{ \begin{pmatrix}1\\1\\\sqrt3+1\\0\end{pmatrix}, \begin{pmatrix}1.572126\\ 1.572126\\ 1.602875 \\ 0.0232641 \end{pmatrix} \right\}.\tag4$$ (see also Wolfram Alpha calculations)

Thus, the issue inequality is satisfied for any non-negative $a,b,c,$ wherein the triple $(a,b,c) = (1,1,1+\sqrt3)$ transforms the inequality to the equality.

$\color{brown}{\textbf{Proved.}}$

Answer 6

This inequality can actually be handled by standard calculus methods. It's a lot of work, but it does work. Just set up the function $$f(x,y,z) =\sqrt{x^3+2}+\sqrt{y^3+2}+\sqrt{z^3+2}-\sqrt{\frac{9+3\sqrt{3}}{2}(x^2+y^2+z^2)}$$ and search for its minimum on the region $(x,y,z)\in [0,\infty)^3$.

First, we look for any critical points in the interior of that region. The gradient is $$\nabla f(x,y,z) = \left(\frac{3x^2}{2\sqrt{x^3+2}}-\sqrt{\frac{9+3\sqrt{3}}{2}}\cdot\frac{2x}{2\sqrt{x^2+y^2+z^2}},\cdots,\cdots\right)$$ where the $\cdots$ represent the cyclically shifted expressions with $y$ and then $z$ in the places $x$ sticks out. Clearing the denominators, this first coordinate of the gradient is zero when $$\sqrt{2}\cdot \sqrt{9+3\sqrt{3}}\cdot x\sqrt{x^3+2}=3x^2\sqrt{x^2+y^2+z^2}$$ $$\frac{2x^3+4}{x^2}=\frac{9}{9+3\sqrt{3}}\left(x^2+y^2+z^2\right)$$ Dividing by $x^2$ as we did doesn't lose any potential solutions since we're looking specifically for critical points in the interior $x,y,z>0$. Now, the right side of that is symmetric in $x,y,z$ and the left side is simply a function of $x$. Applying this to the other two coordinates, we get $$\frac{9}{9+3\sqrt{3}}\left(x^2+y^2+z^2\right) = 2x+\frac{4}{x^2} = 2y+\frac{4}{y^2} = 2z+\frac{4}{z^2}$$ The function $q(t)=2t+\frac{4}{t^2}$ is, unfortunately for us, not monotone. It has a minimum at $\sqrt[3]{4}$, and increases in both directions away from that. Thus, for a given value $k>3\sqrt[3]{4}$, there are two solutions to $q(t)=k$ to choose values of $x,y,z$ from if $x^2+y^2+z^2$ is fixed. We will count points based on how many of $x,y,z$ are on each side of $\sqrt[3]{4}$.

If all three of $x,y,z$ are $\le \sqrt[3]{4}$ or all three are $\le \sqrt[3]{4}$, then $x=y=z$ and $x$ is a root of the polynomial equation $\frac{9}{3+\sqrt{3}}x^4-2x^3-4=0$. That polynomial has exactly one positive root $r$, at about $r\approx 1.582$. Evaluating there, we find $f(r,r,r)\approx 0.0233>0$.

If exactly two of $x,y,z$ are on one side of $\sqrt[3]{4}$, suppose WLOG those are $x$ and $y$. Then, solving the equation $2x+\frac{4}{x^2}=2z+\frac{4}{z^2}$, we find $z=\frac{1+\sqrt{2x^3+1}}{x^2}$. Substituting this in, we seek solutions to $$\frac{3}{3+\sqrt{3}}\left(2x^2+z^2\right)=2x+\frac{4}{x^2}$$ $$\frac{3}{3+\sqrt{3}}\left(2x^2+\frac{1+2x^3+1+2\sqrt{2x^3+1}}{x^4}\right)=2x+\frac{4}{x^2}$$ $$\frac{3}{3+\sqrt{3}}\left(2x^6+2x^3+2+2\sqrt{2x^3+1}\right)=2x^5+4x^2$$ $$\frac{3}{3+\sqrt{3}}\left(u^2+u+1+\sqrt{2u+1}\right)=(u+2)\cdot u^{\frac23}$$ setting $u=x^3$ for convenience in the last equation. Looking for solutions numerically, I started the iteration at $1$... and that was exactly a root. Testing $f$ at $(1,1,1+\sqrt{3})$, it's exactly zero. We have found the equality case.

Numerical calculations find another root at $u\approx 3.886$, for a triple $(x,y,z)\approx (1.572,1.572,1.603)$ and $f(x,y,z)\approx 0.0233$. That's not enough precision to tell whether it's better or worse than the symmetric critical point, but I've got more digits in my spreadsheet. To six significant digits, $f$ is $0.0232638$ at the symmetric critical point and $0.0232641$ at the nearby asymmetric critical points. The symmetric point is slightly better.

And that's it; no more interior critical points. In particular, there are no interior critical points with two or more of $x,y,z$ greater than $\sqrt[3]{4}$.

Part 2

Now we look at the boundary of the region. The two-dimensional boundary has three faces, each of which has exactly one of the $x,y,z$ zero. WLOG, look at the face $x>0,y>0,z=0$.

Here, we eliminate $z$ and look at the two-dimensional gradient of $g(x,y)=f(x,y,0)$. The gradient is almost the same: $$\nabla g(x,y) = \left(\frac{3x^2}{2\sqrt{x^3+2}}-\sqrt{\frac{9+3\sqrt{3}}{2}}\cdot\frac{2x}{2\sqrt{x^2+y^2}},\cdots\right)$$ This is zero when $$\frac{9}{9+3\sqrt{3}}\left(x^2+y^2\right) = 2x+\frac{4}{x^2} = 2y+\frac{4}{y^2}$$ again not needing any truly new calculation. Again, we have two options.

If $x=y$, then $x$ is a root of the polynomial equation $\frac{6}{3+\sqrt{3}}x^4-2x^3-4=0$. That has exactly one positive root, at $x\approx 1.982$ for $g(x,x)\approx 0.2030$.

Alternatively, if $x\neq y$, then $y=\frac{1+\sqrt{2x^3+1}}{x^2}$ and we seek solutions to $$\frac{3}{3+\sqrt{3}}\left(x^2+y^2\right)=2x+\frac{4}{x^2}$$ $$\frac{3}{3+\sqrt{3}}\left(x^2+\frac{1+2x^3+1+2\sqrt{2x^3+1}}{x^4}\right)=2x+\frac{4}{x^2}$$ $$\frac{3}{3+\sqrt{3}}\left(x^6+2x^3+2+2\sqrt{2x^3+1}\right)=2x^5+4x^2$$ $$\frac{3}{3+\sqrt{3}}\left(\frac12u^2+u+1+\sqrt{2u+1}\right)=(u+2)\cdot u^{\frac23}$$ Numerical methods find two roots, at $u\approx 0.756$ and at $u\approx 29.70$. They lead to $x\approx 0.915$ and $x\approx 3.097$, with $y$ being the other root in each case. These critical points have $g(x,y)\approx 0.1044$.

That's the two-dimensional boundary. Now, we look at the one-dimensional boundary, where two of $x,y,z$ are zero. WLOG, let it be the ray $x>0,y=z=0$. Let $h(x)=f(x,0,0)$, so $$h'(x)=\frac{3x^2}{2\sqrt{x^3+2}}-\sqrt{\frac{9+3\sqrt{3}}{2}}$$ This is zero when $$\frac{9}{9+3\sqrt{3}}\cdot x^2 = 2x+\frac{4}{x^2}$$ $$\frac{3}{3+\sqrt{3}}\cdot x^4 - 2x^3-4 = 0$$ This has exactly one root, at $x\approx 3.326$. At that point, $h(x)\approx 0.1956$.

Finally, the one-dimensional boundary $x=y=z=0$. At that point, $f(0,0,0)=3\sqrt{2}\approx 4.2426$.

Also, we must consider what happens as any of $x,y,z$ go to $\infty$. Suppose WLOG that $x\ge y\ge z\ge 0$. Then $f(x,y,z) < \sqrt{x^3}-\sqrt{\frac{9+3\sqrt{3}}{2}(x^2+x^2+x^2)} = x\left(\sqrt{x}-\sqrt{\frac{9+3\sqrt{3}}{2}}\right)$, which goes to $\infty$ as $x\to\infty$. The entirety of the infinite boundary is the supremum for our function.

Coda

Summarizing, the complete list of critical points:
$(x,y,z)\approx (1.582,1.582,1.582)$ for $f(x,y,z)\approx 0.0233$
$(x,y,z)\approx (1.572,1.572,1.603)$ and permutations for $f(x,y,z)\approx 0.0233$
$(x,y,z) = (1,1,1+\sqrt{3})$ and permutations for $f(x,y,z)=0$
$(x,y,z)\approx (1.982,1.982,0)$ and permutations for $f(x,y,z)\approx 0.2030$
$(x,y,z)\approx (0.915,3.097,0)$ and permutations for $f(x,y,z)\approx 0.1044$
$(x,y,z)\approx (3.326,0,0)$ and permutations for $f(x,y,z)\approx 0.1956$
$(x,y,z) = (0,0,0)$ for $f(x,y,z) = 3\sqrt{2}$
As $\max(x,y,z)\to\infty$, $f(x,y,z)\to\infty$

The smallest value of $f$ at any critical point is $0$ at $(1,1,1+\sqrt{3})$. This is therefore the global minimum of $f$, and we have the inequality $f(x,y,z)\ge 0$ for all $x,y,z$. That is the inequality we wanted to prove, and we're done.