MHB Can you prove the inequality challenge?

AI Thread Summary
The inequality challenge requires proving that for any real number x greater than or equal to 1/2 and any positive integer n, the expression x^(2n) is greater than or equal to (x-1)^(2n) + (2x-1)^n. The discussion highlights the use of the binomial expansion to establish that (a+b)^n is greater than or equal to a^n + b^n when a and b are positive. By substituting a = 2x - 1 and b = (x - 1)^2, the inequality simplifies to x^(2n) being greater than or equal to (2x - 1)^n + (x - 1)^(2n). This approach effectively demonstrates the validity of the original inequality. The conversation concludes with acknowledgment of the contributions made by participants.
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Let $x\ge \dfrac{1}{2}$ be a real number and $n$ a positive integer. Prove that $x^{2n}\ge (x-1)^{2n}+(2x-1)^n$.
 
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anemone said:
Let $x\ge \dfrac{1}{2}$ be a real number and $n$ a positive integer. Prove that $x^{2n}\ge (x-1)^{2n}+(2x-1)^n$.
[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]
 
Well done, Opalg(Yes) and thanks for participating!:)
 
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