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.
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