<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] 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.
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