# Finding all solutions for Bernoulli inequality

in our course we have been given the task to find all real solutions to the inequality

$(1+x)^n \geq 1 + nx$ ($\forall n \in \mathbb{N}$).

I have already proved this for $x \geq -1$ using induction and worked out the behavior for $-2 \leq x < -1$.

What I have trouble with is proving the fact, that the inequality does not hold for $x \leq -2$. Some other posts I have read about this problem used derivatives, which are not yet in my repertoire...

I hope some of you can point me to the right direction, thanks in advance.

• For $\;x=-2\;,\;\;n=1\;$ it is already false. For some values of $\;n\;$ though the inequality still holds, can you see? – DonAntonio Nov 27 '16 at 12:12
• Ty for your answer, I think I see your point. I corrected my post above, cause the inequality should hold for all $n \in \mathbb{N}$ – JoRa Nov 27 '16 at 12:48
• see the following for your proof:en.wikipedia.org/wiki/Bernoulli's_inequality – Dr. Sonnhard Graubner Nov 27 '16 at 12:52

I assume you want those $x$ so that it holds for all $n$, $1+nx\leq (1+x)^n$.

In that case one can show by induction that for $x\leq -2$ that $$(1+x)^{2n+1}< 1+(2n+1)x.$$

I would be glad to see your solution for the interval $(-2,-1)$.

Consider the case $x \in [-2, 0]$: Now we want to show that $f(x) = (1+x)^n - (1+nx)$ that f is strictly monotonically decreasing on this interval.

Let $y < z$ : $$$$\begin{split} f(z) - f(y) &= ((1+z)^n - (1+nz)) - ((1+y)^n - (1+ny)) \\ &= (1+z)^n - (1+y)^n + n(y-z) \\ &= (z-y)(\sum_{k=0}^{n-1}(1+z)^k(1+y)^{n-k} - n) \end{split}$$$$

Where $\sum_{k=0}^{n-1}(1+z)^k(1+y)^{n-k} < n$ since $|1+z|, |1+y| \le 1$ but $1+z = 1+y = 1$ not possible under the requirement $y < z$.

It follows, that $f(z) - f(y) < 0$ $\implies$ strictly monotonically decreasing. Since $f(0)=0$ the Bernoulli-inequality hold for this interval.

Consider the last case $x < -2$: Define a = -2 - x $\implies$ $1+x = -(1+a)$

For odd n you can find: $$$$\begin{split} (1+nx) - (1+x)^n &= (-1)^n ((1+x)^n - (1+nx)) \\ &= (1+a)^n - (-1)^n (1-n(a+2)) \\ &\ge 1 + na + \frac{n(n-1)}{2}a^2 \\ &\ge 1+ na + \frac{n(n-1)}{2}a^2 + 1 - n(a+2) \\ &= (n-1)(\frac{n}{2}a^2 - 2) \\ &= (n-1)\frac{na^2-4}{2} \\ &> a \end{split}$$$$ for $n > \frac{4}{a^2} = \frac{4}{(x+2)^2}$, so that the inequality does not hold