About contradictory inequality

Question

We know that if $a,b,c≥0$

$a^3+b^3+c^3≥3abc$

Let $abc=1$

We get: $$a^3+b^3+c^3≥3$$

Then,

$a^3+b^3+c^3+(3abc)≥4\sqrt[4]{3}abc$

And let $abc=1$

We get:

$$a^3+b^3+c^3≥4\sqrt[4]{3}-3$$

Update:

For example: $a^3+b^3+c^3=4\sqrt[4]{3}-3+k$

Here,

if $0<k<6-4\sqrt[4]{3}$

We get, $a^3+b^3+c^3<3$

A Contradiction?! What is wrong?

Answer 1

Because in your AM-GM the equality occurs for the equality case of all variables,

but $3abc=a^3=b^3=c^3$ for positive variables gives

that in your last inequality the equality does not occur and it's indeed so.

We have no any contradiction.

If you want to use AM-GM for the second expression you can write the following. $$a^3+b^3+c^3+3abc\geq6\sqrt[6]{a^3b^3c^3(abc)^3}=6abc,$$ which is very good, it's just Osem.

Answer 2

You should evaluate $a^3,b^3,and\ c^3$ when $a=b=c=1$ to get a smallest possible value for the left side expression (because AM-GM inequality becomes equal when a=b=c, and in this case we have abc=1), and see that the inequality is alright.

$3\ge3$ and $3\ge\sqrt[4]{3}-3$ have no contradictions.

Answer 3

We do not have any contradiction here. One of the equation gives $a^3+b^3+c^3\ge 3$ and the other equation gives $a^3+b^3+c^3\ge ~2.2649$.

It would have been a contradiction if the signs were opposite.