If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.

Question

If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.

I recently came across a question in which we had to prove the above inequality using the given condition as mentioned above. Here $a,b,c$ are distinct positive integers and $k$ is also a positive integer. I absolutely have got no idea how to solve it or efficiently use the condition 'positive integers'. Furthermore, although the expression $a^3+b^3+c^3-3abc$ seems a bit familiar but I'm not able to understand how to make the condition useful.

Please help.

Answer 1

From $\displaystyle a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\bigg[(a-b)^2+(b-c)^2+(c-a)^2\bigg]\geq 3.$

Because $a,b,c$ are distinct integers.

So here we have taken $a,b,c$ as $3$ consecutive integers.

And $\displaystyle (a+b+c)^2-3(ab+bc+ca)\geq 0\Rightarrow (a+b+c)^2\geq 3(ab+bc+ca)=3(3k^2-1)$

So $\displaystyle (a+b+c)\geq \sqrt{9k^2-3}$

So we have taken $(a+b+c)\geq 3k.$

So we have $\displaystyle a^3+b^3+c^3-3abc\geq 3\cdot (3k)=9k.$

Equality hold when $a,b,c$ are $3$ consicutive posotive integers

Namely we have taken $k-1,k,k+1$

Answer 2

Use that $$a^3+b^3+c^3-3abc=\left( a+b+c \right) \left( {a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2} \right) $$