# How to prove that $1- \frac{x^2}{n} \leq (1+\frac{x}{n})^n\cdotp(1-\frac{x}{n})^n$

How would I prove this inequality (assuming its true, its from a textbook)

$$1 - \frac{x^2}{n} \leq (1+\frac{x}{n})^n\cdotp(1+\frac{-x}{n})^n$$

if $n > |x|$, $x\in R$ and $n\in N$

I first rewrote the inequality to $$1 - \frac{x^2}{n} \leq (1-\frac{x^2}{n^2})^n$$I then tried to manipulate the inequalities by saying the right hand side was greater than a smaller expression however I was unable to prove the above. I also tried induction where the base case works however I was unable to show that a case being true implies the next also being true.

Any help would be appreciated

• Good first step. Now Bernoulli's inequality would help. – Daniel Fischer Jul 28 '14 at 14:01
• How would you use the Bernoulli inequality? – bob Jul 28 '14 at 14:05
• What does the Bernoulli inequality say? How can you relate that to the expression $\left(1-\frac{x^2}{n^2}\right)^n$? – Daniel Fischer Jul 28 '14 at 14:07
• So, $(1-\frac{x^2}{n^2})^n > 1 - n\frac{x^2}{n^2} = 1-\frac{x^2}{n}$ ? – bob Jul 28 '14 at 14:11
• Observe that the function $$f (x)=(1-\frac{x^2}{n^2})^2-(1-\frac {x^2}{n})$$ is increasing for all $|x|\ge n$. – Mohammad W. Alomari Jul 28 '14 at 14:13

$$1-\frac{x^2}{n}\overset{?}{\leq}\left(1-\frac{x^2}{n^2}\right)^n$$ is a good starting point. You can assume $|x|<\sqrt{n}$, since otherwise the inequality is trivial, with the LHS being non positive and the RHS being positive.

Consider the logarithm of both sides. Then: $$\log\left(1-\frac{x^2}{n}\right)\leq n\log\left(1-\frac{x^2}{n^2}\right)$$ is a consequence of the inequality: $$\forall z\in[0,1),\qquad \log(1-z)\leq n\log\left(1-\frac{z}{n}\right)$$ that follows from the fact that: $$\int_{0}^{z}\frac{dx}{1-x}\geq\int_{0}^{z}\frac{dx}{1-\frac{x}{n}}$$ since $(1-x)\leq 1-\frac{x}{n}$.

• Sorry, I can't understand why the last inequality is a Jensen's inequality. – Shine Jul 28 '14 at 14:16
• @Shine: I modified my argument in order to avoid it, I think it is more elementary now. – Jack D'Aurizio Jul 28 '14 at 14:35

Deriving both sides on $x$, $$-\frac{2x}{n}\le-n\frac{2x}{n^2}(1-x^2)^{n-1},$$ or $$-1\le-(1-\frac{x^2}{n^2})^{n-1}.$$ The latter relation is obviously true for $|x|<n$, so that the LHS of the initial relation decreases faster than the RHS, while they are equal for $x=0$.

(If you prefer, $l'(x)\le r'(x)\implies l'(x)-r'(x)\le0\implies l(x)-r(x)$ is decreasing $\implies l(x)-r(x)\le l(0)-r(0)=0\implies l(x)\le r(x)$.)

• No you cannot differentiate the "inequality" (not equality) this is not true mathematically even your solution correct or not. – Mohammad W. Alomari Jul 28 '14 at 14:22
• What if $C^1$ continuity is guaranteed ? – Yves Daoust Jul 28 '14 at 14:28
• No. Example: it is well know that $\log (1+x)<x$ for all $x>-1$, $x\ne 0$. If you differentiate both sides we get $\frac {1}{1+x}< 1$ for all $x>-1$ which is not true take $x=-1/2$ – Mohammad W. Alomari Jul 28 '14 at 14:34
• The logics goes the other way: $\frac1{1+x}>1$ for $x<0$, so that $\log(1+x)-x$ increases until $0$, where it equals $0$. Hence, $\log(1+x)-x$ is negative and $\log(1+x)<x$. I am not saying $f'(x)>g'(x)\implies f(x)>g(x)$, I am saying $f'(x)>g'(x)\implies f(x)-g(x)$ is increasing. – Yves Daoust Jul 28 '14 at 14:47
• You wrote $\frac{1}{x+1}>1$ for $x <0$, this is not true take $x =-1-\frac {1}{3}$ – Mohammad W. Alomari Jul 28 '14 at 15:45