The inequality challenge states that for any real number \( x \ge \frac{1}{2} \) and positive integer \( n \), the inequality \( x^{2n} \ge (x-1)^{2n} + (2x-1)^n \) holds true. The proof utilizes the binomial expansion, demonstrating that \( (a+b)^n \geq a^n + b^n \) for positive \( a \) and \( b \). By setting \( a = 2x-1 \) and \( b = (x-1)^2 \), the inequality is transformed into a valid form, confirming the original statement. This proof was effectively summarized by the participant Opalg.
PREREQUISITESMathematicians, students studying advanced algebra, and anyone interested in the field of inequalities and mathematical proofs will benefit from this discussion.
[sp]If $a$ and $b$ are positive then $(a+b)^n \geqslant a^n+b^n$ (because the binomial expansion of the left side consists of the two terms on the right side, together with other terms which are all positive). Put $a=2x-1$ and $b = (x-1)^2$. Then $a+b = x^2$ and the inequality becomes $x^{2n} \geqslant (2x-1)^n + (x-1)^{2n}.$ [/sp]anemone said:Let $x\ge \dfrac{1}{2}$ be a real number and $n$ a positive integer. Prove that $x^{2n}\ge (x-1)^{2n}+(2x-1)^n$.