How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$

This answer is neither nice nor complete. At the end, I still need to plot a very complicated function in one variable to conclude the inequality is true. In any event, here is my attempt.

Let $a,b,c$ be the three elementary symmetric polynomials associated with $x,y,z$, i.e: $$ \begin{cases} a &= x + y + z\\ b &= xy + yz + zx\\ c &= xyz \end{cases} \quad\iff\quad (\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - a\lambda^2 + b\lambda - c $$ In terms of $a,b,c$, the inequality we want to prove can be rewritten as: $$\begin{align} & c + \sqrt{b^2 - 2ac} -\frac43 \sqrt{ac} \stackrel{?}{\ge} 0\tag{*1}\\ \iff & ( \sqrt{c} - \frac23\sqrt{a} )^2 + \sqrt{b^2 - 2ac} -\frac{4a}{9} \stackrel{?}{\ge} 0\tag{*2} \end{align}$$ and the condition $x^2 + y^2 + z^2 = 1$ is equivalent to $a^2 = 1 + 2b$.

Over the domain of our problem $$\mathscr{D} = \{ (x,y,z) : x, y, z \ge 0, x^2 + y^2 + z^2 = 1 \},$$ we have: $$\begin{align} & 0 \le b = xy + yz + zx \le x^2 + y^2 + z^2 = 1\\ & 1 \le a = \sqrt{1+2b} \le \sqrt{3}\\ & 0 \le c = xyz = ( (x^2y^2z^2)^{\frac13} )^{\frac32} \le ((x^2+y^2+z^2)/3)^{\frac32} = \frac{1}{\sqrt{27}}. \end{align}$$ Since both sides of above inequalities are reachable at $(1,0,0)$ and at $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$,
the admissible ranges for $a,b,c$ over $\mathscr{D}$ are $[0,1]$, $[1,\sqrt{3}]$ and $[0, \frac{1}{\sqrt{27}}]$ respectively.

Notice over $\mathscr{D}$, $$\sqrt{c} \le \frac{1}{\sqrt[4]{27}} < \frac23 \le \frac23 \sqrt{a}$$ In the L.H.S of $(*2)$, if we fix $b$ ( and hence $a = \sqrt{1+2b}$ ) and consider it as a function of $c$, it is a decreasing function. This implies in order to prove the inequality $(*1 \iff *2)$, we only need to verify $(*1)$ when $c$ attains its maximum admissible value $c_{max}(b)$ subject to given $b$:

$$c_{max}(b) := \sup \{\; xyz : (x,y,z) \in \mathscr{D}, xy + yz + zx = b\;\}$$

Let us rewrite $c$ as a function of $b$ and $t = \min\{x,y,z\}$, we have: $$\begin{align} c = & xyz = (b - (a - t)t)t = (b - at + t^2)t \\ = & (t-\frac{a}{3})^3 - \frac{a^2-3b}{3} t + \frac{a^3}{27}\\ = & (t-\frac{a}{3})^3 - \frac{d^2}{3}(t - \frac{a}{3}) + \frac{a^3 - 3ad^2}{27} \end{align}$$ where $d^2 = a^2 - 3b = 1-b$. Notice $\frac{\partial c}{\partial t} = 3 (t - \frac{a}{3})^2 - \frac{d^2}{3}$. One can verify $c$ achieves its maximum at $t = \frac{a - d}{3}$. As a result, we get:

$$\begin{align} & c_{max}(b) = -\frac{d^3}{27} + \frac{d^3}{9} + \frac{a^3 - 3ad^2}{27} = \frac{2 d^3 + a^3 - 3ad^2}{27} = \frac{(a+2d)(a-d)^2}{27}\\ \implies & a c_{max}(b) = \frac{( (a+d)^2 - d^2 )(a - d)^2}{27} = \frac{b^2}{3}( 1 - L(b) ) \end{align}$$ where $$L(b) = \left( \frac{d}{a+d} \right)^2 = \frac{1-b}{(\sqrt{1+2b}+\sqrt{1-b})^2}$$ In terms of $b$ and $L(b)$, the inequality $(*1)$ at $c = c_{max}(b)$ becomes:

$$ \frac{b^2}{3\sqrt{1+2b}}(1 - L(b)) + \sqrt{\frac{b^2}{3}(1 + 2L(b))} - \frac43 \sqrt{\frac{b^2}{3}( 1 - L(b) )} \stackrel{?}{\ge} 0\tag{*3} $$

I don't know how to simplify this further but the plot below show that it is true for all $b \in [0,1]$.

So the inequality at hand is indeed true. In fact the inequality is strict except when $$\begin{cases} b = 0, &\iff (x,y,z) = (1,0,0) \text{ or } (0,1,0) \text{ or } (0,0,1)\\ b = 1, &\iff (x,y,z) = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ) \end{cases}$$

We need to prove that $$xyz+\sqrt{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)}\geq\frac{4}{3}\sqrt{xyz(x+y+z)(x^2+y^2+z^2)}.$$

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=w^3+3\sqrt{(3u^2-2v^2)(3v^4-2uw^3)}-4\sqrt{u(3u^2-2v^2)w^3}.$$ But by AM-GM $$f'(w^3)=1-\frac{3u\sqrt{3u^2-2v^2}}{\sqrt{3v^4-2uw^3}}-2\sqrt{\frac{u(3u^2-2v^2)}{w^3}}\leq1-\sqrt{\frac{u(3u^2-2v^2)}{w^3}}\leq0,$$ which says that $f$ is a decreasing function and it's enough to prove our inequality

for a maximal value of $w^3$, which happens for equality case of two variables.

Since our inequality is homogeneous, it's enough to do it for $y=z=1$, which gives $$x+\sqrt{(x^2+2)(2x^2+1)}\geq\frac{4}{3}\sqrt{x(x+2)(x^2+2)}$$ or after squaring of the both sides $$x^4-16x^3+11x^2-32x+9+9x\sqrt{(x^2+2)(2x^2+1)}\geq0.$$ Now, by C-S $$\sqrt{(x^2+2)(2x^2+1)}=\sqrt{(x^2+1+1)(x^2+x^2+1)}\geq x^2+x+1.$$ Id est, it remains to prove that $$x^4-16x^3+11x^2-32x+9+9x(x^2+x+1)\geq0$$ or $$(x-1)^2(x^2-5x+9)\geq0.$$ Done!