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!
 A: 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.
A: Let $$f(a,b,c,\lambda)=abc-1+\sqrt{\frac{2}{3}}(a-c)+\lambda(a^2+b^2+c^2-3).$$
Thus, in the minimum point we need
$$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial\lambda}=0,$$
which gives:
$$bc+\sqrt{\frac{2}{3}}+2\lambda a=ac+2\lambda b=ab-\sqrt{\frac{2}{3}}+2\lambda c=0.$$
Now, if $c=0$, so $$3=a^2+b^2\leq2a^2,$$ which gives $$a\geq\sqrt{\frac{3}{2}}$$ and $$abc-1+\sqrt{\frac{2}{3}}(a-c)=\sqrt{\frac{2}{3}}a-1\geq0.$$
Now, let $c>0$.
Thus, $$ \frac{bc+\sqrt{\frac{2}{3}}}{a}=\frac{ab-\sqrt{\frac{2}{3}}}{c}=\frac{ac}{b},$$ which gives
$$b^2c+b\sqrt{\frac{2}{3}}=a^2c$$ and $$b^2a-b\sqrt{\frac{2}{3}}=ac^2,$$ which after summing gives $$b^2=ac,$$ which with our condition gives $$a^2+ac+c^2=3$$ and we need to prove that
$$\sqrt{a^3c^3}+\sqrt{\frac{2}{3}}(a-c)\cdot\frac{a^2+ac+c^2}{3}\geq\sqrt{\left(\frac{a^2+ac+c^2}{3}\right)^3}.$$
Now, let $a=xc$ and $a^2+c^2=2uac.$
Thus, $x\geq1$ and $u\geq1$ and we need to prove that:
$$\sqrt{x^3}+\sqrt{\frac{2(x^2+1-2x)}{3}}\cdot\frac{x^2+x+1}{3}\geq\sqrt{\left(\frac{x^2+x+1}{3}\right)^3}$$ or
$$1+\sqrt{\frac{4(u-1)}{3}}\cdot\frac{2u+1}{3}\geq\sqrt{\left(\frac{2u+1}{3}\right)^3}$$ or
$$27+4(u-1)(2u+1)^2+12\sqrt{3(u-1)}(2u+1)\geq(2u+1)^3$$ or
$$(u-1)(4u^2-2u-11)+6\sqrt{3(u-1)}(2u+1)\geq0,$$ which is obvious for $4u^2-2u-11>0$ or $u>\frac{1+\sqrt{45}}{4}.$
Id est, it's enough to prove that $$6\sqrt{3}(2u+1)\geq\sqrt{u-1}(-4u^2+2u+11)$$ for $$1\leq u\leq \frac{1+\sqrt{45}}{4}.$$
Indeed, we need to prove that:
$$108(2u+1)^2\geq(u-1)(4u^2-2u-11)^2$$ or
$$229+355u+304u^2+68u^3+32u^4-16u^5\geq0$$ or
$$229+355u+304u^2+24u^3+24u^4+4u^3(11+2u-4u^2)\geq0$$ and we are done in this case.
Also, we need to check, what happens for $b=c$ and for $a=b$.
Two these cases lead to inequalities of one variable.
I hope there is a solution without LM.
A: 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 $f(a)\ge1$.
