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Let $x_1,...,x_m$ be non-negative real numbers. Is the following inequality correct? $$\sum_{n=1}^{m}x_n^2\geq \left(\sum_{n=1}^{m}\dfrac{n}{2^n}x_n\right)^2.$$

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  • $\begingroup$ This sounds like a job for the QM-AM inequality. $\endgroup$
    – Magma
    Commented Feb 10, 2020 at 18:27

1 Answer 1

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It is true. Using Cauchy-Schwarz:

$$\left(\sum_{n=1}^{m}\dfrac{n}{2^n}x_n\right)^2 \leq \left(\sum_{n=1}^{m}\dfrac{n^2}{4^n}\right)\left(\sum_{n=1}^mx_n^2\right)$$

and we can prove that:

$$\sum_{n=1}^{m}\dfrac{n^2}{4^n}=\frac{5\cdot 4^{m+1}-9m^2-24m-20}{27\cdot 4^m} < 1$$

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  • $\begingroup$ Thanks for your answer $\endgroup$ Commented Feb 11, 2020 at 3:44

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