# g and f : ℝ -> ℝ, $f(x)=x^2; g(x)=x$. Prove that f(x) ≥g(x), for all x≥1

I tried proving this by contradiction and so I proved $$f(x) does not hold for all x≥1.

I just took an example of x, eg x=1. $$f(1), the statement is false. Therefore, proof complete.

I would only like to know if my proof is correct and if is not I would like some suggestions about dealing with my problem, thank you.

• Seems unnecessarily complicated to put this as a function analysis terms. It's simply stating for all $x \ge 1$ that $x^2 \ge x$. That's simply a matter of noting if $1 \le x$ then $1\times x \le x\times x$. That's all. Commented Feb 14, 2019 at 1:27
• @fleablood does my proof make sense even though is complicated? Thanks, I totally understand what you are saying. Commented Feb 14, 2019 at 1:30
• "I just took an example of x" One example doesn't prove anything! Consider trying to prove all cats are black and trying to to a proof by contradiction. I'll assume no cats are black. I pick up Felix the Cat. He's black! Contradiction: proof done. All cats are black. Commented Feb 14, 2019 at 1:33
• The negation is that there is some $x \ge 1$ where $f(x) > g(x)$. $x =1$ is not it. To do a proof by contradiction you have to find that no possible $x$ will do. Not just $1$. Commented Feb 14, 2019 at 1:37
• As flea says, $f(1)=1\not<1=g(1)$. Commented Feb 14, 2019 at 1:39

I'll critique your proof. You're trying to refute the theorem by finding a counterexample. With that correction, so far, so good -- that's a perfectly valid technique. But testing one example and discovering that it's not a counterexample is not, of course, a proof.

Your theorem is: $$\forall x ~(x \geq 1 \Rightarrow x^2 \geq x$$). So to find a counterexample you need to find an $$x$$ so that this implication fails. That means you need an example of $$x$$ such that $$x \geq 1$$ and $$x^2 \lt x$$. Note that this second inequality has to be strict.

This is where your attempt fails. You tried to use $$x=1$$ as your counterexample, but $$1^2 = 1$$ so you don't have strict inequality, and that means you don't have a counterexample.

In fact, for the reasons discussed above, there are no counterexamples because the theorem is true.

If $$1 \le x$$ then $$0 < x$$ and $$1\cdot x \le x\cdot x$$.

That's all.

Finding one instance which works does not, as pointed out by others, in anyway constitute a proof. That only serves to disprove the claim that something is never true.

In terms of proving it, we can employ a similar method to the one you suggest, but make our example arbitrary. Let's say that $$x=1+t; t> 0$$

Then: $$(1+t)^2 =1+2t+t^2$$

Undeniably $$2t>t$$ for $$t>0$$, therefore $$(1+t)^2>1+t$$

Equality is at $$t=0$$ and this is easy to show.