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We are already familiar with Bernoulli's inequality: $(1+px)\le(1+x)^p$ for $x\ge-1, p\ge1$.

Can this be generalized to say something useful about $(1+p_1x_1+\cdots+p_nx_n)$? For instance, one would hope that something like this could hold: $$(1+p_1x_1+\cdots+p_nx_n) \le (1+x_1+\cdots+x_n)^{p_1+\cdots+p_n}$$ However, this doesn't seem obvious. If you have any other ideas than using Bernoulli, that's great too, of course.

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up vote 2 down vote accepted

$$(1+p_1x_1+\cdots+p_nx_n) \leq (1+\frac{p_1x_1+\cdots+p_nx_n}{p_1+...+p_n})^{p_1+...+p_n} \leq (1+x_1+\cdots+x_n)^{p_1+...+p_n}\,.$$


$$p_1x_1+\cdots+p_nx_n < (p_1+..+p_n)(x_1+..x_n) \,.$$

This is true for example if $x_1,..x_n \geq 0$, but also in many other cases.

Negative $x_i$ I n general the inequality is not true if we allow for negative $x_i$, the problem is that $p_1,..,p_n$ are independent.

For example, pick $x_1>0, x_2 <0$ so that $x_1+x_2 <0$. Then for all $p_1, p_2$ we have

$$(1+x_1+x_2)^{p_1+p_2} <1 \,.$$

but, if $p_1 > \frac{-p_2x_2}{x_1}$ then

$$(1+p_1x_1+p_2x_2)>1 \,.$$

Note that $x_1+x_2<0$ can be made as close to 0 as one wants.

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Is it possible to obtain the generalization for negative $x_i$, such as $x_i\ge -1$? How did you obtain your results? – cont-math Aug 2 '12 at 16:22
Nope, check my addon. – N. S. Aug 2 '12 at 17:00

Even better: at least for nonnegative $x_i$ and $p_i$, $$ 1+p_1x_1+\cdots p_nx_n\leqslant(1+x_1)^{p_1}\cdots(1+x_n)^{p_n}, $$ and $$ (1+x_1)^{p_1}\cdots(1+x_n)^{p_n}\leqslant(1+x_1+\cdots+x_n)^{p_1+\cdots+p_n}. $$

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Does this not work for negative $x_i$, such that $x_i\ge -1$ continues to work for each $i$? How did you obtain your results? – cont-math Aug 2 '12 at 16:21
Let me recommend at least a dose of skepticism about the result when $x_1+\cdots+x_n=-1$ and $p_1x_1+\cdots+p_nx_n\gt-1$. // To obtain the inequality in the post (and probably a slight extension of the conditions), try a recursion on $n$. – Did Aug 2 '12 at 16:26
@did you are right, see the addition to my answer. – N. S. Aug 2 '12 at 17:01
@N.S. Sure. You might have noted that my previous comment does provide an explicit counterexample, valid as soon as $n\geqslant3$. – Did Aug 2 '12 at 17:16
I'm not seeing how recursion, or even induction, can help with this. – cont-math Aug 2 '12 at 21:08

We can use the binomial theorem to prove that $ (1+x)^p \geq ( 1 + p x )\,,$ $p\geq 1$ (finite natural number), $x \geq 0\,, $ that is

$$ (1+x)^p = 1 + p x + \frac{p(p-1)}{2} x^2 + \dots+x^p \geq 1 + p x \,,$$

for $ x\geq 0\,, p \geq 1\,. $

Now apply the binomial theorem to $ ( 1 + x_1 + x_2 )^{p_1+p_2} $ for $x \geq 0\,, p_i \geq 1\,, i=1,2 \,, $ ($p_1\,,p_2$ are finite natural numbers)

$$(1+x_1+x_2)^{p_1+p_2} = 1+(p_1+p_2)(x_1+x_2)+\dots+(x_1+x_2)^{p_1+p_2} ) = 1 + p_1 x_1 + p_2 x_2 + p_1 x_2 + p_2 x_1 + \dots \geq 1 + p_1 x_1 + p_2 x_2 \,, $$ for $ x_i\geq 0\,, p_i \geq 1 \,,i=1\,,2 \,. $

One can generalize this process to the general case in your problem. I'd like you to investigate the case when $ x_i \geq -1\,, i = 1,2,\dots, n\,.$

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