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I've got most of the inductive work done but I'm stuck near the very end. I'm not so great with using induction when inequalities are involved, so I have no idea how to get what I need...

\begin{align} (1+x)^{k+1}&\geq (1+x)\left[1+kx+\frac{k(k-1)}{2}x^2\right]\\ &=1+kx+\frac{k(k-1)}{2}x^2+x+kx^2+\frac{k(k-1)}{2}x^3\\ &=1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3 \end{align}

And here's where I have no clue how to continue. I thought of factoring out $kx^2$ from the remaining three terms, but I don't see how that can get me anywhere.

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  • $\begingroup$ Use binomial theorem. $\endgroup$ – Hanul Jeon Sep 28 '13 at 3:45
  • $\begingroup$ I need to use induction... I suppose I should have mentioned that. $\endgroup$ – agent154 Sep 28 '13 at 3:46
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    $\begingroup$ $k+\frac{k(k-1)}{2}=\frac{(k+1)k}{2}$ $\endgroup$ – wj32 Sep 28 '13 at 3:49
  • $\begingroup$ @wj32 Aha I didn't see that. Thanks. $\endgroup$ – agent154 Sep 28 '13 at 3:55
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We need to show $(1+x)^{k+1}\ge1+(k+1)x+\frac{(k+1)k}2x^2$

You already have $$ \begin{align} (1+x)^{k+1}&\geq1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3 \end{align}$$ $$=1+(k+1)x+x^2\frac{k(k+1)}2+\frac{k(k-1)}{2}x^3$$

which is $$\ge1+(k+1)x+x^2\frac{k(k+1)}2$$ if $\displaystyle\frac{k(k-1)}{2}x^3\ge0$ which is true for $x\ge0$ and $k\ge1$

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If $n=1$, it is trivial. Suppose it is true for $n$. We will show that this formula is true for $n+1$. $$ \begin{aligned} (1+x)^{n+1}&=(1+x)^n(1+x)\\ &\ge \left( 1+nx+\frac{n(n-1)}{2} x^2 \right)(1+x)\\ &= 1+nx+\frac{n(n-1)}{2}x^2 +x+nx^2 +\frac{n(n-1)}{2} x^3\\ &= 1+(n+1)x+\frac{n^2-n+2n}{2}x^2+\frac{n(n-1)}{2} x^3\\ &= 1+(n+1)x +\frac{n^2+n}{2}x^2+\frac{n(n-1)}{2} x^3\\ &\ge 1+(n+1)x +\frac{n(n+1)}{2}x^2\\ &= 1+(n+1)x +\frac{(n+1)(n+1-1)}{2}x^2 \end{aligned} $$ so we get desired result.

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\begin{align*} &= 1 + (k + 1)x + \frac{k(k+1)}{2}x^2 + \frac{k(k-1)}{2}x^3\\ &\geq 1 + (k + 1)x + \frac{k(k+1)}{2}x^2\\ \end{align*}

That's the inductive step. You're done.

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