# Use a direct proof to show that if $x\gt 1$ then $x^5 +x+1\gt 2$.

Use a direct proof to show that if $x\gt 1$ then $x^5 +x+1\gt 2$.

It's obvious that if $x\gt 1$ then $x+1\gt 2$ , also $x^5\gt0 \;as \;x\gt1\gt0$, thus $x^5+x+1\gt 2$.

Is it fine ?

I can't understand what do they mean by direct proof?

• Yes, that proof is fine. Commented Feb 16, 2016 at 21:27
• Is this a direct proof ?
– User
Commented Feb 16, 2016 at 21:28
• Directly we get $x^5+x+1>1^5+1+1=3$, which is better. Commented Feb 16, 2016 at 21:29
• I suppose by "direct" proof a proof which isn't a proof by contradiction is meant. Your proof clearly isn't by contradiction, so should be fine. Commented Feb 16, 2016 at 21:29
• Yes, it is direct in the sense that you are showing that it is true without resorting to any roundabout arguments, or contradiction, or induction, or various theorems, etc. Commented Feb 16, 2016 at 21:29

Since the derivative $f'(x)=5x^4+1$ is always positive, the function $f(x)=x^5+x+1$ is increasing over $\mathbb{R}$ and $x>1$ implies $f(x)>f(1)=3$.
• The value at $1$ is $3$, but otherwise, yes. Commented Aug 11, 2016 at 1:53