 # Use derivatives to prove that if n\geq1, then (1+x)^{n}>1+nx for -1<x<0 and 0<x (notice that equality holds for x=0). rocedwrp 2021-07-03 Answered

Use derivatives to prove that if $$\displaystyle{n}\geq{1}$$, then $$\displaystyle{\left({1}+{x}\right)}^{{{n}}}{>}{1}+{n}{x}$$ for -1

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Let $$\displaystyle{n}\geq{1}$$. Then we have to show that $$\displaystyle{\left({1}+{x}\right)}{\left\lbrace{n}\right\rbrace}{>}{1}+{n}{s}$$, for -10.
Let us consider the function $$\displaystyle{g{{\left({x}\right)}}}={\left({1}+{x}\right)}^{{{n}}}—{\left({1}+{n}{x}\right)}$$. Then we note that $$g(0) = 0.$$ Moreover, by calculating the derivative we see that
$$\displaystyle{g}'{\left({x}\right)}={n}{\left({1}+{x}\right)}^{{{n}-{1}}}-{n}={n}{\left({\left({1}+{x}\right)}^{{{n}-{1}}}—{1}\right)}$$.
Now since $$\displaystyle{n}-{1}\geq{1}$$ we see that for $$x>0$$,
$$\displaystyle{\left({1}+{x}\right)}^{{{n}-{1}}}-{1}{>}{0}$$,
that is, $$g'(x)>0$$ for $$x>0$$. On the other hand, if -1 $$\displaystyle{\left({1}+{x}\right)}^{{{n}-{1}}}{<}{1}$$
which then shows that $$g'(x)<0$$ on -10 and is decreasing on -10 on $$\displaystyle{\left({0},\infty\right)}$$ and $$g(x)<0$$ on -1 $$\displaystyle{\left({1}+{x}\right)}^{{{n}}}{>}{1}+{n}{x}$$, for -10.
Thus, it completes the proof.
Let $$\displaystyle{n}\geq{1}$$. Then $$(1+x)^{n}>1+nx,$$ for -10