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!

Heres another way.  First note:

*

*From the expression, it is enough to consider the case $c \geqslant \max(a, b)$

*From homogeneity, we may set $c=1$.  So $a, b \in [0, 1]$.

*Replacing $a, b $ with their arithmetic mean reduces the numerator of LHS, leaving the denominator untouched, as $a^3+b^3, a^2+b^2, -ab$ all become smaller.  Hence we it is enough to consider $a=b=t$.

Finally we are left to show for $t \in [0, 1]$:
$$\frac{2t^3+2t(t-1)-5t^2}{(2t+1)^3} \geqslant -\frac15 \iff \frac{(3t-1)^2}{5(2t+1)^2}\geqslant 0$$
