..........................

WakeUp wrote: Let $t\ge\frac{1}{2}$ be a real number and $n$ a positive integer. Prove that \[t^{2n}\ge (t-1)^{2n}+(2t-1)^n\] By Minkowski's inequality, \begin{align*} \sqrt[n]{(t-1)^{2n} + (2t-1)^n} &= \sqrt[n]{(t^2 - 2t + 1)^n + (2t - 1)^n} \\ &\leq \sqrt[n]{(t^2 - 2t + 1)^n} + \sqrt[n]{(2t-1)^n} \\ &\leq |(t-1)^2| + |2t - 1| \\ &= t^2 - 2t + 1 + 2t - 1 \\ &= t^2. \end{align*} Raising both sides to the $n$th power yields the desired result.

The inequality can be rewritten as $$((t-1)^2+(2t-1))^n\geq(t-1)^{2n}+(2t-1)^n$$which is true since $(x+y)^n\geq x^n+y^n$ for any non-negative $x$ and $y$.