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How to prove that $(1+x)^{2n}>1+2nx$ , $x\neq 0$ using induction?

Any hint would be appreciated.

felipeuni
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    http://www.proofwiki.org/wiki/Bernoulli's_Inequality or http://math.stackexchange.com/questions/204196/proof-by-induction-nn1-n-1n-1-xn-ge-1-nx-other-inequal or http://science.kennesaw.edu/~plaval/math4381/induction.pdf – lab bhattacharjee Mar 11 '14 at 05:45

4 Answers4

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If $x>0$, by the induction hypothesis: $$(1+x)^{2n}=(1+x)^2(1+x)^{2n-2}>(1+x)^2(1+2nx - 2x)$$ $$=1+2nx + ((2 n -2)x^3 + (4 n - 3)x^2)>1+2nx$$ Q.E.D.

Nathaniel Bubis
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Here is my way. Let $f(x)=(1+x)^{2n}-2nx$ on $[0,\infty)$.You can easily show that $f$ is increasing.(see what happens to $f'$??).Therefore for $t>0$ we obtain $1=f(0)<f(t)=(1+t)^{2n}-2nt$ or equivalently $(1+t)^{2n}>1+2nt$.

Similar way you can show that inequality holds for $x<0$

BigM
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Base case: $$(1+x)^{2\cdot1}=1+2x+x^2>1+2\cdot1\cdot x$$ Induction Hypothesis:

Assume that $$(1+x)^{2k}>1+2kx$$ Induction step: $$ \begin{align} (1+x)^{2(k+1)}&=(1+x)^{2k}\cdot(1+x)^2 \\ &>(1+x)^{2k}\cdot(1+2x) \tag{By the Base Case} \\ &>(1+2kx)(1+2x) \tag{By the Induction Hypothesis} \\ &=1+2x+2kx+4kx^2 \\ &>1+2x+2kx \\ &>1+2x(k+1) \\ &\text{QED} \end{align} $$

Vibhav
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    Note that $1+2x$ is not necessarily positive. – felipeuni Mar 11 '14 at 07:25
  • Well, then $(1+2kx)$ and $(1+2x)$ would both be negative, which would then be multiplied in Step 3 of the Induction Step, (making the expression positive) after which we would get back to Step 4.

    Also, you could try starting with $(1+x)^{2k}>|1+2kx|$ (though I haven't tried that out).

     – Vibhav Mar 11 '14 at 10:38
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The 2 years old answers are somewhat incomplete regarding the negative case so here we go: $f(x)=(1+x)^{2n}$ with $n\geq 1$ has as second derivative: $$ f''(x) = 2n (2n-1) (1+x)^{2(n-1)} $$ which is strictly positive for $x\neq 0$. $f$ is therefore strictly convex whence strictly above its tangents except at the touching point. And $f(0)+f'(0)x = 1 + 2n x$ is the tangent at $(0,1)$.

H. H. Rugh
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  • The question asks how to do it "using induction". Therefore, this is not an answer. – GEdgar Aug 20 '16 at 23:10
  • Hmmm, you are right. Sort of got caught by a catch 22. I responded to a new post with the same title which did not ask for induction but which was blocked as a duplicate of the present. It was suggested that I add an answer to this one... (alternatively one could of course repair one of the above inductive answers)... – H. H. Rugh Aug 20 '16 at 23:24