An unusual inequality problem involving real exponential power Problem.

Let $\alpha\in\mathbb R,\, \omega \in\mathbb R_{+}$ and $n\in\mathbb Z_{\geq 5}$, then prove that
$$4\omega ^{\alpha }+\omega^{n-4\alpha}-n\omega+n\geq 5.$$

The source of the problem is a student in a Telegram group.  He claims that the problem can be solved using only algebra. Using the derivative and induction is restricted here.  The author claims that this inequality can be proved without using derivatives and mathematical induction. The problem is exactly as I wrote.
My attempts:
I think I have proven this inequality for $n=5.$
Let $n=5$, then we have
$$4\omega ^{\alpha }+\omega^{5-4\alpha}-5\omega\geq 0\\ 4\omega ^{\alpha-1 }+\omega^{4-4\alpha}-5\geq 0.\\4\omega ^{\alpha-1 }+\frac 1{\omega^{4\alpha-4}}-5\geq 0.
$$
$\omega^{\alpha-1}=t,t>0$
$$4t+\frac {1}{t^4}-5\geq 0\\
4t^5-5t^4+1\geq 0$$
$t=1$ is double root. By polynomial long division, we have
$$4t^5-5t^4+1=(t - 1)^2 (4 t^3 + 3 t^2 + 2 t + 1)\geq 0$$
Equality occurs when $t=1\implies \alpha =1 \,\, \text{or}\,\,\omega=1$.
Then I tried to prove the inequality for $n=6$, I got
$$4\omega ^{\alpha }+\omega^{6-4\alpha}-6\omega+1\geq 0.$$
I see that $\omega=1$ is the root of the equation. But, this doesn't help.
Also, I realized that if $\omega =1$, then
$$4+1-n+n≥5\implies 5\geq 5$$
This implies, for $\omega=1$, we obtain an equality condition.
I couldn't try anything more.  Because power is a real number, not an integer.  Therefore, I haven't been able to make much progress.
 A: The form $4\omega^\alpha+\omega^{n-4\alpha}$ strongly suggests it's AGM time:
$$
\frac{4\omega^\alpha+\omega^{n-4\alpha}}{5}\ge\sqrt[5]{\omega^{4\alpha}\omega^{n-4\alpha}} = \omega^{n/5}.
$$
So we have
$$
4\omega^\alpha+\omega^{n-4\alpha}-n\omega+n \ge 5\omega^{n/5} -n(\omega-1)
$$
Next we want to use Bernoulli's inequality $x^r \ge 1+r(x-1)$ for $r\ge 1$ and $x\ge 0$ to get $\omega^{n/5}\ge 1+(n/5)(x-1)$. However, in case your friend claims that's not algebraic even though it is for rational exponents, we can also get it from another application of AGM:
$$
\frac{5\omega^{n/5} + (n-5)}{n}\ge \sqrt[n]{\omega^n 1^{n-5}} = \omega\Longrightarrow 5\omega^{n/5}\ge n(\omega-1)+5.
$$
Putting this into the inequality above gives
$$
4\omega^\alpha+\omega^{n-4\alpha}-n\omega+n \ge 5\omega^{n/5} -n(\omega-1)\ge  n(\omega-1)+5-n(\omega-1)  = 5.
$$
Of course, this could also all be done in one step by just rearranging the inequality to
$$
\frac{4\omega^\alpha + \omega^{n-4\alpha} +n-5}{n} \ge \sqrt[n]{\omega^{4\alpha}\omega^{n-4\alpha}1^{n-5}} = \omega,
$$
which is just AGM directly, but I think the two step process is more clear on how you would come up with the answer.
