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anemone
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Let $x\ge \dfrac{1}{2}$ be a real number and $n$ a positive integer. Prove that $x^{2n}\ge (x-1)^{2n}+(2x-1)^n$.
[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 $x^{2n}\ge (x-1)^{2n}+(2x-1)^n$.
"Inequality Challenge IX" is a yearly event organized by a group of researchers and scientists to address issues related to inequality in various aspects of society.
The purpose of "Inequality Challenge IX" is to bring together experts from different fields to discuss and find solutions for the growing problem of inequality in the world.
The theme for "Inequality Challenge IX" is chosen based on current global trends and issues related to inequality. It is decided by a committee of experts and researchers.
"Inequality Challenge IX" is open to anyone interested in the topic of inequality, including scientists, researchers, policymakers, and the general public.
There are several ways to get involved with "Inequality Challenge IX", such as submitting a research paper, attending the event, or volunteering to help with organizing. You can also follow the event on social media and spread awareness about the issue of inequality.