<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] 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 $f(a) \geqslant 0.$
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