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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?

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    $\begingroup$ Yes, that proof is fine. $\endgroup$ – Théophile Feb 16 '16 at 21:27
  • $\begingroup$ Is this a direct proof ? $\endgroup$ – User Feb 16 '16 at 21:28
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    $\begingroup$ Directly we get $x^5+x+1>1^5+1+1=3$, which is better. $\endgroup$ – Dietrich Burde Feb 16 '16 at 21:29
  • $\begingroup$ 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. $\endgroup$ – Wojowu Feb 16 '16 at 21:29
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    $\begingroup$ 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. $\endgroup$ – Théophile Feb 16 '16 at 21:29
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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 $f(x)>f(1)=3$.

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  • $\begingroup$ Dat moment, when even using the derivative is a sledgehammer proof (see Dietrich's comment) but +1 I guess $\endgroup$ – Patrick Da Silva Aug 11 '16 at 0:52
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    $\begingroup$ If you want to be gentler you can use the increasingness of addition and multiplication. This is learned before the libidinal impulse. $\endgroup$ – Jacob Wakem Aug 11 '16 at 1:36
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The function is monotone increasing for positive values. This is because addition and multiplication are increasing in either argument. Moreover the value at 1 is 3. The theorem follows.

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    $\begingroup$ The value at $1$ is $3$, but otherwise, yes. $\endgroup$ – Brian Tung Aug 11 '16 at 1:53
  • $\begingroup$ @BrianTung do you know the Grothendieck prime? $\endgroup$ – Jacob Wakem Aug 11 '16 at 3:30
  • $\begingroup$ I might have heard that story before? I don't remember. Anyway, a good one. :-) $\endgroup$ – Brian Tung Aug 11 '16 at 16:07

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