# Inequality with a floor function

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.

• Use floor(x) > x-1. Plug it into your inequality. – Benjamin Wang 2 days ago

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 $$x \geq 4$$