Proving $\frac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\frac{1}{5}$ For $a,b,c\geqslant 0.$ Prove$:$
$$\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\dfrac{1}{5}$$
I found an AM-GM proof.
Since $$P+\frac{1}{5}\geqslant 0\Leftrightarrow 6\,{a}^{3}+6\,{b}^{3}+8\,{a}^{2}c-2\,a{c}^{2}+8\,{b}^{2}c-2\,b{c}^{2}-19\,abc+3\,{a}^{2}b+3\,a{b}^{2}+{c}^{3} \geqslant 0$$
And by AM-GM$:$
$$2\,a{c}^{2}\leqslant 6{a}^{3}+\frac49{c}^{3},$$
$$2\,b{c}^{2}\leqslant 6{b }^{3}+\frac49{c}^{3},$$
$$19\,abc\leqslant \frac19{c}^{3}+3a{b}^{2}+3{a}^{2}b+8 \,{a}^{2}c+8\,{b}^{2}c.$$
So we are done!
Is there another nice proof$?$ Thanks for a real lot!

For the fun : with algebra.
Consider that you look for the minimum value of function
$$f=\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3}$$ with $c=1$ as
@Macavity explained. Computing the derivatives and simplify, we end with the two equations
$$3 a (a+4) b+2 a (a+2)-3 b^3-8 b^2-3 b-1=0\tag1$$
$$-3 a^3-8 a^2+3 a (b (b+4)-1)+2 b (b+2)-1=0\tag2$$ Use $(2)$ to compute $b$ (it is just a quadratic). So
$$b=\frac{\sqrt{9 a^4+30 a^3+61 a^2+33 a+6}-6 a-2}{3 a+2}$$ Pluf in $(1)$ and get a monster. But after a few suarin steps, the only acceptable solution is $a=\frac 13$ so $b=\frac 13$ too and $f_{min}=-\frac 15$.
