Is the following inequality true $(a^3-b^6)^3+(3abc)^3 \leq (a^3-b^6+3cb^3)^3$?

Question

Let $a,b,c$ be all positive integers greater than $1$.

If $$a>b^2$$ and $$a^3-b^6> 3c$$ Is this the following inequality true?: $$(a^3-b^6)^3+(3abc)^3 \leq (a^3-b^6+3cb^3)^3$$ I tried to expand without much success. Any Hints?

Answer 1

No, it is not true. There are infinitely many counterexamples.

For $(a,b,c)=(k^3,k,k^5)\ (k\ge 2\in\mathbb Z)$, one has $$a-b^2=k^3-k^2=k^2(k-1)\gt 0$$ and $$a^3-b^6-3c=k^9-k^6-3k^5=k^5(k^4-k-3)\ge k^5(2^4-2-3)\gt 0,$$ but $$(a^3-b^6)^3+(3abc)^3-(a^3-b^6+3cb^3)^3$$ $$=(k^9-k^6)^3+(3k^9)^3-(k^9-k^6+3k^8)^3$$ $$=9k^{20}(k-1)^2(k^2+k+1)(3k^3+(k-1)(k+1)+k(k-1))\color{red}{\gt} 0$$

Answer 2

For $a=10$, $b=3$ and $c=89$, we have: $$a=10\quad >\quad b^2=9$$ $$a^3-b^6=271\quad >\quad 3c=267$$ But: $$(a^3-b^6)^3+(3abc)^3=513,942,303,511 \not\le (a^3-b^6+3cb^3)^3=418,508,992,000$$ Hence the inequality is false.

Answer 3

It seems the following.

I can simplify the inequality by the following equivalent transformations:

$$(a^3-b^6)^3+(3abc)^3 \leq (a^3-b^6+3cb^3)^3$$

Put $A=a^3-b^6$.

$$A^3+(3abc)^3 \leq (A+3cb^3)^3$$

$$(3abc)^3 \leq (A+3cb^3)^3-A^3$$ $$(3abc)^3 \leq (3cb^3)((A+3cb^3)^2+(A+3cb^3)A+A^2) $$ $$9a^3c^2 \leq (A+3cb^3)^2+(A+3cb^3)A+A^2$$ $$9a^3c^2 \leq A^2+6Acb^3+9c^2b^6+A^2+3Acb^3+A^2$$ $$9a^3c^2 \leq 3A^2+9Acb^3+9c^2b^6$$ $$3a^3c^2 \leq A^2+3Acb^3+3c^2b^6$$ $$0 \leq A^2+3Acb^3+3c^2(b^6-a^3)$$ $$0 \leq A^2+3Acb^3-3c^2A$$ Since $A>0$, we can divide by $A$: $$ 0 \leq A+3cb^3-3c^2$$ $$ 0 \leq a^3-b^6+3cb^3-3c^2.$$