Proving $abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$ The question is this:

If $a\ge b\ge c\ge 0$ and $a^2+b^2+c^2=3$, then prove that
$$abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$$

For my work on this inequality, I have proved already under constraints that it is true.
Proof for: $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0.$
$$
\sqrt{3}abc + 
\sqrt{2}a  -
\sqrt{3} -
\sqrt{2}c 
\geqslant 0
$$
$$
a\left(
\sqrt{3}bc + 
\sqrt{2}
\right) 
+ (-1)\left(
\sqrt{3} + 
\sqrt{2}c 
\right) \geqslant 0
$$
$$
(1 + 1)(a\left(
\sqrt{3}bc + 
\sqrt{2}
\right) 
+ (-1)\left(
\sqrt{3} + 
\sqrt{2}c 
\right)) \geqslant 0
$$
By Chebyshev,
$$
(a - 1)
(\sqrt{3}bc + 
\sqrt{2} + 
\sqrt{3} +  
\sqrt{2}c 
)\geqslant0
$$
$$
a \geqslant 1
$$
Chebyshev Inequality requires the sequences to be monotonous. As $a+1>0$, we need to have the other sequence also in the same order, hence the condition: $\sqrt{3}bc + \sqrt{2} \geqslant\sqrt{3} + \sqrt{2}c$. The sequences are $(a,-1)$ and $(\sqrt{3}bc + \sqrt{2} ,\sqrt{3} + \sqrt{2}c)$.
I have tried another way but that was untrue. I have reached this far. The constraint $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0$ isn't true always. Try $(a,b,c) = (\sqrt{3},0,0)$.
Thanks for extensions or other solutions too are welcome!

Let $f(a,b)=abc+(a-c)k$ where $c^2=3-a^2-b^2$ and $k=\sqrt{2/3}$. Assuming $c\ne0$, $$f_a=bc+(1-c_a)k=0$$ for critical points. Now $c_a=-a/c$ so $bc^2+(c+a)k=0$, contradicting $c\ne0$. This means that either $c=0$, or any other solutions must lie on the boundaries of the constraints, which are:

*

*$a=b$ which yields $f(a)=(a^2-k)\sqrt{3-2a^2}+ak$;


*$b=c$ which yields $f(a)=a(3-a^2)/2+(a-\sqrt{(3-a^2)/2})k$.
When $c=0$ we have $a^2+b^2=3$ such that $a\ge b$ so $a\ge\sqrt{3/2}$ and $f(a,b)=0+ak\ge1$.
For the first case we have $a\ge c\implies a\ge1$ so the domain of $f$ is $[1,\sqrt{3/2}]$. Notice that $f(1)=f(\sqrt{3/2})=1$ and $f(a)-1$ is positive. Similarly, for the second case we also have $a\ge1$. Note that $f(1)=1$ and calculus yields $f(a)\ge1$.
