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I'm trying to prove that $8x^2-5\text {floor} (x)^2 \ge 10$ for all $x \ge 4$? I understand that I can use $0 \le x - \text {floor} (x) < 1$ to assist in solving the equation but I'm having trouble fitting it in.

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  • $\begingroup$ Use floor(x) > x-1. Plug it into your inequality. $\endgroup$ – Benjamin Wang 2 days ago
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we know $8x^2 \geq 10 + 5x^2$ for all $x \geq 4$... Just solve the inequality or draw graph///

and $x^2 \geq \text{floor(x)}^2$ for all $x \geq 0$... Obviously floor of x will be less or equal to x when x is postive.

Hence $8x^2 \geq 10 + 5x^2\geq 10 + 5\text{floor(x)}^2$ for all $x \geq 4$

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