<s>[INST] 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!
 [/INST] 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 $$\left(x-1\right) \left[ (x-y)(2x+2y+11)+3(y-1)(x+y+3) \right] \\
= (x-1)\left[2x^2+8x+3xy + y^2 - 5y - 9\right] > (x-1) (y-1)^2 > 0 \, .$$
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