Maybe Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$a^3+b^3+c^3\geq abc+2 $$

sqing wrote: Maybe Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$a^3+b^3+c^3\geq abc+2 $$ With my method works also for this inequality too.

Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$\frac{a^3+b^3+c^3}{3}\geq\frac{abc+k-1}{k} $$Where $k\in N ^+.$

sqing wrote: Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$a^3+b^3+c^3\geq abc+2 $$

dangerousliri wrote: sqing wrote: Maybe Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$a^3+b^3+c^3\geq abc+2 $$ With my method works also for this inequality too. Looking forward to seeing your proof. Thanks.

sqing wrote: dangerousliri wrote: sqing wrote: Maybe Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$a^3+b^3+c^3\geq abc+2 $$ With my method works also for this inequality too. Looking forward to seeing your proof. Thanks. It is kind of similar to your solution. If $abc\geq 1$ then $a^3+b^3+c^3\geq 3abc\geq abc+2$. If $abc<1$ then $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3\geq abc(a+b+c)+3\Rightarrow (1-abc)(a+b+c-3)\geq 0\Rightarrow a+b+c\geq 3$, hence $a^3+b^3+c^3\geq \frac{(a+b+c)^3}{9}\geq 3>abc+2$.

dangerousliri wrote: Let $a,b$ and $c$ be positive real numbers such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that the following inequality hold, $$\frac{a^3+b^3+c^3}{3}\geq\frac{abc+2021}{2022}.$$

sqing wrote: Let $a,b$ and $c$ be positive reals such that $a+b+c+3abc\geq (ab)^2+(bc)^2+(ca)^2+3$. Show that $$\frac{a^3+b^3+c^3}{3}\geq\frac{abc+k-1}{k} $$Where $k\in N ^+.$