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

Question

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

I just took an example of x, eg x=1. $f(1)<g(1)$, 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.

Answer 1

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

That's all.

Answer 2

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.

Answer 3

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.