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!
 A: We write inequality as
$$a^3+8(b+c)a^2+(8b^2-19bc-2c^2)a+b^3+8b^2c-2bc^2+c^3 \geqslant 0.$$
Because $a^3-ab(2a-b) = a(a-b)^2 \geqslant 0,$ so we will show that
$$ab(2a-b)+8(b+c)a^2+(8b^2-19bc-2c^2)a+b^3+8b^2c-2bc^2+c^3 \geqslant 0,$$
or
$$f(a) = 2(5b+4c)a^2+(7b^2-19bc-2c^2)a+b^3+8b^2c-2bc^2+c^3 \geqslant 0.$$
Because $b^3+8b^2c-2bc^2+c^3 \geqslant 0,$ therefore
If $7b^2-19bc-2c^2 \geqslant 0$ then $f(a) \geqslant 0.$
If $7b^2-19bc-2c^2 \leqslant 0,$ we have
$$\Delta_a = (b^2-68bc-28c^2)(3b-c)^2 \leqslant 0.$$
So $f(a) \geqslant 0.$
A: 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 $t \in [0, 1]$:
$$\frac{2t^3+2t(t-1)-5t^2}{(2t+1)^3} \geqslant -\frac15 \iff \frac{(3t-1)^2}{5(2t+1)^2}\geqslant 0$$
A: 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 $f_{min}=-\frac 15$.
A: The LHS is homogenous of degree $0$. Hence we can replace $a=\frac{cx}{3}$ and $b=\frac{cy}{3}$ with $x,y\geq 0$ and obtain the equivalent objective $$2x^3 + x^2y + xy^2 + 2y^3 + 8x^2 - 19xy + 8y^2 - 6x - 6y + 9 \geq 0 \\
\text{or} \\ (x-y)^2 \left(2x+2y+11\right) + 3(x-1)(y-1)\left(x+y+3\right) \geq 0 \, .$$
The last equation is symmetric with respect to $x$ and $y$ and can only be negative if the second term is negative, that is when WLOG $x > 1$ and $y < 1$. In this case the LHS is larger than $$\left(x-1\right) \left[ (x-y)(2x+2y+11)+3(y-1)(x+y+3) \right] \\
= (x-1)\left[2x^2+8x+3xy + y^2 - 5y - 9\right] > (x-1) (y-1)^2 > 0 \, .$$
