<s>[INST] 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!
 [/INST] 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.
</s>