Let x_k be real numbers such that x_k geq k^4 | vedclass

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(A) Let $n = 2018$. We are given $N = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.For inequality $I$,we use the Cauchy-Schwarz inequality in the form $(\sum

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both $I$ and $II$ are true

Vedclass · Answer

(A) Let $n = 2018$. We are given $N = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.For inequality $I$,we use the Cauchy-Schwarz inequality in the form $(\sum a_k b_k)^2 \leq (\sum a_k^2)(\sum b_k^2)$.Let $a_k = \sqrt{k}$ and $b_k = \sqrt{k} x_k$. Then:$\left(\sum_{k=1}^{n} \sqrt{k} \cdot \sqrt{k} x_k\right)^2 \leq \left(\sum_{k=1}^{n} (\sqrt{k})^2\right) \left(\sum_{k=1}^{n} (\sqrt{k} x_k)^2\right)$$\left(\sum_{k=1}^{n} k x_k\right)^2 \leq \left(\sum_{k=1}^{n} k\right) \left(\sum_{k=1}^{n} k x_k^2\right) = N \sum_{k=1}^{n} k x_k^2$.Thus,$I$ is true.For inequality $II$,we note that since $k \geq 1$,we have $k^2 \geq k$ for all $k \in \{1, 2, \dots, n\}$.Therefore,$\sum_{k=1}^{n} k^2 x_k^2 \geq \sum_{k=1}^{n} k x_k^2$.Since $I$ is true,we have $\left(\sum_{k=1}^{n} k x_k\right)^2 \leq N \sum_{k=1}^{n} k x_k^2 \leq N \sum_{k=1}^{n} k^2 x_k^2$.Thus,$II$ is also true.Therefore,both $I$ and $II$ are true.

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