Prove that $n(a^2 - a + 1)^n \ge a^{2n} + a^{2n(1 - \frac{1}{n - 1})} + \cdots + a^{\frac{2n}{n - 1}} + 1$.

Question

Given $a \ge 1$ and $n \ge 2$, prove that $$\large n(a^2 - a + 1)^n \ge a^{2n} + a^{2n\left(1 - \frac{1}{n - 1}\right)} + \cdots + a^{\frac{2n}{n - 1}} + 1$$

I've tried using the Chebyshev inequality for multiple powers of $a$ but it didn't work. The solution may involve basic elements of calculus, at which I suck. So please help me understand your problem understandably (obviously) because I am still weak at calculus if your solution includes calculus.

Answer 1

For $n=4$ let $a=x^3$.

Thus, we need to prove that $$4(x^6-x^3+1)^4\geq x^{24}+x^{16}+x^8+1.$$ Are you sure that it's true?

Try $x=1.1.$

For $n=3$ your inequality is true: $$3(a^2-a+1)^3-(a^6+a^3+1)=(a-1)^4(2a^2-a+2)\geq0.$$