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Prove the following statement using induction.

If $x > 1$ is a real number, then $(x+1)^n > nx^2+1$ for all $n \ge 3$.

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Math Lover
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JBreal1
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    Ways to improve your question: 1) Ask questions one at a time, 2) Make the effort to write the question here rather than simply posting a link to a picture, 3) Don't just post homework here without any thoughts/efforts of your own, 4) Write a more informative title. Your last question also get heavily downvoted and closed because of this. – John Doe Dec 01 '17 at 01:26
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    Welcome to stackexchange. You are more likely to get answers rather than downvotes and votes to close if you edit your question so that there is just one question (not 8), included as text (not an image) and if you show what you have tried and where you are stuck. – Ethan Bolker Dec 01 '17 at 01:27
  • I forgot to write down which one when I realized it I couldn’t change it – JBreal1 Dec 01 '17 at 01:28
  • @josebravo You can edit the question. Please also include some text about where you are stuck. – John Doe Dec 01 '17 at 01:29
  • How do I edit ? I’m new to this – JBreal1 Dec 01 '17 at 01:30
  • There is a button beneath the question, but above the comments. It is in line with where your name is. – John Doe Dec 01 '17 at 01:31
  • Thank you for your feedback – JBreal1 Dec 01 '17 at 01:33
  • Try to first do it to $n=3$. Then suppose it is valid for $n$ and do it for $n = n+3$. Then you should find a way to simplify it and get your answer. – A.T Dec 01 '17 at 01:37
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    To replace that photograph with just the question you meant to ask, in a legible mathematical format, start with the advice here: https://math.stackexchange.com/help/notation – David K Dec 01 '17 at 01:38
  • Side note: l liked the problem 2! It is a lovely problem to which induction can be applied as well. –  Dec 01 '17 at 01:58

3 Answers3

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I leave the base case $n=3$ to you.

Note that you need to prove that $$(x+1)^n>nx^2+1 .$$ Suppose $$(x+1)^k >kx^2+1$$ for some $k \ge 3$. Then $$(x+1)^{k+1}=(x+1)(x+1)^k >(x+1)(kx^2+1)=kx^3+x+kx^2+1.$$ Next, we consider $kx^3+x > x^2 \iff kx^2-x+1>0 \iff 1-4k <0$, which is true.

Therefore, $$(x+1)^{k+1}>kx^3+x+kx^2+1>x^2+kx^2+1=(k+1)x^2+1.$$

The expression $ax^2+bx+c > 0$ when $a >0$ and $b^2-4ac < 0$.

Math Lover
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First verify it for $n=3$:

$(x+1)^3 > 3x^2 +1 $

$(x+1)^3 = x^3 + 3x^2 + 3x+1 = x^3+3x +( 3x^2 +1) > 3x^2+1$

$x^3+3x >0$, which is true, since $x>1$.

Suppose it holds for $n$. Now check for $n=n+3$:

$(x+1)^{n+3} > (n+3)x^2 +1 $

$(x+1)^n (x+1)^3 > nx^2+3x^2 +1 $

$\frac{(x+1)^n (x+1)^3}{(x+1)^n } > \frac{nx^2+1}{(x+1)^n } + \frac{3x^2}{(x+1)^n }$

$ (x+1)^3 > \frac{nx^2+1}{(x+1)^n } + \frac{3x^2}{(x+1)^n } $

But, since by hypothesis $(x+1)^n > nx^2 +1 \Rightarrow 1>\frac{nx^2+1}{(x+1)^n }$ and also $3x^2 > \frac{3x^2}{(x+1)^n} $

Thus:

$ (x+1)^3 > 3x^2+1 > \frac{3x^2}{(x+1)^n } + \frac{nx^2+1}{(x+1)^n }$

Which is true, since it is valid for $n=3$

A.T
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$(x + 1)^{n+1} = (x+1)^n*(x+1) > (nx^2 + 1)(x+1) = nx^3 + nx^2 + x + 1$.

$(n+1)x^2 + 1= nx^2 + x^2 + 1$

Which of those two are larger?

fleablood
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