1

Prove $(0.25 + x^2) \ge x \iff x^2-x + \frac 14 \ge 0.$

The solution is pretty clear for $x\ge 0.5$ and $x \le0.25$.

But I was unable to prove the inequality for $0.25 < x < 0.5$.

amWhy
  • 210,739
Rajesh Marndi
  • 469
  • See for example https://math.stackexchange.com/a/2881060/42969 – Martin R Apr 04 '22 at 11:16
  • Moving everything to the left side of the equation, by subtracting x from both sides, we get, $x^2-x+ \frac 14 \geq 0 \iff (x-\frac 12)^2\geq 0$. No matter the value of $x$, what is in the parentheses, when squared, will be $\geq 0$. So the solutions for $x$ include $x\in \mathbb R.$ – amWhy Apr 04 '22 at 15:41

1 Answers1

5

The question is equivalent to proving $$x^2 - x + \frac{1}{4} \geq 0 \iff 4x^2 - 4x + 1 \geq 0 \iff (2x - 1)^2 \geq 0$$