Proving $abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$

If $a\geq b\geq c\geq0$ then prove $$3\sqrt3abc+\sqrt2\left(a-c\right)\left(a^2+b^2+c^2\right)\geq\left(a^2+b^2+c^2\right)^{\frac{3}2}.$$ Case 1: $c=0,$ it's obvious. Equality at $a=b\iff a=b=\sqrt{\frac{3}2}.$

Case 2: $c=1.$ If $a=1,$ then we are done. Equality at $a=b=c=1.$ If $a>1$ then consider on $[1,a]$ the function $$f(b):=3\sqrt3ab+\sqrt2\left(a-1\right)\left(a^2+b^2+1\right)-\left(a^2+b^2+1\right)^{\frac{3}2}.$$ We have: $$f'(b)=b\left(\frac{3\sqrt3a}b+2\sqrt2\left(a-1\right)-3\sqrt{a^2+b^2+1}\right)\implies$$ $f$ is pseudo-concave $\implies\min_{b\in[1,a]}{f(b)}\in\{f(1),f(a)\}.$ But $$f(1)>0$$ and $$f(a)>\sqrt3\left(2a^2+1\right)+\sqrt2\left(a-1\right)\left(2a^2+1\right)-\left(2a^2+1\right)^{\frac{3}2}>0.$$ We are done. Edit: Let me give further details about $f(1)>0.$ We need to prove $$3\sqrt3a+\sqrt2\left(a-1\right)\left(a^2+2\right)>\left(a^2+2\right)^{\frac{3}2}\iff$$ $$6\sqrt6a\left(a-1\right)\left(a^2+2\right)>a\left(a-1\right)^2\left(-a^3+2a^2+a+16\right)\iff$$ $$a^4-3a^3+3a^2+a^2\left(-2+6\sqrt6\right)-15a+16+12\sqrt6>0,$$ which is obviously true.

Denote \begin{align} P &= abc - 1 + \sqrt{\frac{2}{3}}(a-c),\\ Q &= \frac{a^2+b^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(\sqrt{\tfrac{a^2+b^2}{2}} - c). \end{align}

First, it is easy to prove $Q= \frac{3-c^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(\sqrt{\tfrac{3-c^2}{2}} - c) \ge 0$ (note: $c\in [0,1]$). Indeed, if $c\in [0, \frac{1}{2}]$, we have \begin{align} Q &= \frac{3-c^2}{2}c - 1 + \sqrt{1 - \frac{c^2}{3}} - \sqrt{\tfrac{2}{3}}\ c\\ &\ge \frac{3-c^2}{2}c - 1 + 1 - \frac{c^2}{3} - \sqrt{\tfrac{2}{3}}\ c \\ &= \frac{1}{6}c(-3c^2 - 2c + 9 - 2\sqrt{6})\\ &\ge 0, \end{align} and if $c\in (\frac{1}{2}, 1]$, we have \begin{align} Q &= \frac{3-c^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(\sqrt{1 + \tfrac{1-c^2}{2}} - c)\\ &\ge \frac{3-c^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(1 + \tfrac{1}{3}\cdot \tfrac{1-c^2}{2} - c)\\ &= \frac{1}{18}(1-c)[9c^2 + (\sqrt{6} + 9)c + 7\sqrt{6} - 18]\\ &\ge 0. \end{align}

Second, we have (let $x = \frac{b}{a} \in [0, 1]$) \begin{align} &P - Q\\ =\ & (ab - \tfrac{a^2+b^2}{2})c + \sqrt{\tfrac{2}{3}}(a - \sqrt{\tfrac{a^2+b^2}{2}})\\ =\ & \sqrt{\tfrac{2}{3}}\frac{\frac{a^2 - b^2}{2}}{a + \sqrt{\tfrac{a^2+b^2}{2}}} - \frac{(a-b)^2}{2} c\\ =\ & \frac{a-b}{2} \left[\sqrt{\tfrac{2}{3}}\frac{a + b}{a + \sqrt{\tfrac{a^2+b^2}{2}}} - (a-b)c\right]\\ \ge\ & \frac{a-b}{2} \left[\sqrt{\tfrac{2}{3}}\frac{a + b}{a + \sqrt{\tfrac{a^2+b^2}{2}}} - (a-b)b\right]\\ =\ & \frac{a-b}{2}\left[\sqrt{\tfrac{2}{3}}\frac{1 + x}{1 + \sqrt{\tfrac{1+x^2}{2}}} - (1-x)x a^2\right]\\ \ge\ & \frac{a-b}{2}\left[\sqrt{\tfrac{2}{3}}\frac{1 + x}{1 + \sqrt{\tfrac{1+x^2}{2}}} - (1-x)x \frac{3}{1+x^2}\right]\\ \ge\ & \frac{a-b}{2}\left[\sqrt{\tfrac{2}{3}}\frac{1 + x}{1 + 1 - \frac{1-x^2}{4}} - (1-x)x \frac{3}{1+x^2}\right]\\ \ge\ & \frac{a-b}{2}\cdot \frac{9x^4 + (4\sqrt{6}-9)x^3 + (4\sqrt{6}+63)x^2 + (4\sqrt{6}-63)x+4\sqrt{6}}{3(x^2+7)(x^2+1)}\\ \ge\ & \frac{a-b}{2}\cdot \frac{(4\sqrt{6}+63)x^2 + (4\sqrt{6}-63)x+4\sqrt{6}}{3(x^2+7)(x^2+1)}\\ \ge\ & 0 \end{align} where we have used $3 = a^2 + b^2 + c^2 \ge a^2 + a^2x^2$ to obtain $a^2 \le \frac{3}{1+x^2}$, and we have used $\sqrt{\tfrac{1+x^2}{2}} = \sqrt{1 - \frac{1-x^2}{2}} \le 1 - \frac{1-x^2}{4} $.

We are done.