The inequality m - x^2 <= 1 - x has a positive solution. What values can m take?

Step-by-step solution

1. 1

   Isolate the parameter rather than the variable. Because the question is about $m$, solve the inequality for $m$:

   $m-x^{2}\le 1-x\ \Longleftrightarrow\ m\le x^{2}-x+1$

   Now the condition reads: there exists $x>0$ with $m\le f(x)$, where $f(x)=x^{2}-x+1$.

2. 2

   Translate 'there exists' into a bound on the range. A value $m$ works iff $m$ is less than or equal to some value of $f$ on $x>0$ — that is, iff

   $m\le \sup_{x>0}f(x)$

   This is the step that decides the whole problem, and it is where 'has a solution' (existential) must not be confused with 'holds for all $x>0$' (universal). The universal version would instead require $m\le\min_{x>0}f(x)$.

3. 3

   Analyse $f$ on the positive axis. Completing the square,

   $f(x)=x^{2}-x+1=\left(x-\frac12\right)^{2}+\frac34$

   so $f$ has minimum $\tfrac34$ at $x=\tfrac12$ (which lies in $x>0$), and $f(x)\to+\infty$ as $x\to+\infty$. Its range on $x>0$ is therefore $\left[\tfrac34,+\infty\right)$ — unbounded above.

4. 4

   Apply the criterion. Since $\sup_{x>0}f(x)=+\infty$, the condition $m\le f(x)$ can always be met: whatever $m$ is, choose $x$ large enough that $x^{2}-x+1\ge m$.

   $m\in\mathbb{R}$

5. 5

   Exhibit a witness to make it concrete. Given any $m$, take $x=\max(1,m)$. Then $f(x)=x^{2}-x+1=x(x-1)+1\ge 1$ and $f(x)\ge x\ge m$ whenever $x\ge 1$, so that $x$ is a positive solution. For example $m=100$ is satisfied by $x=100$, since $100^{2}-100+1=9901\ge 100$.

6. 6

   Contrast with the 'for all' version. Had the problem said the inequality holds for every $x>0$, the answer would have been $m\le\tfrac34$ instead — the minimum, not the supremum. Reading the quantifier correctly is the entire difficulty of this question.