Use induction to prove $\prod_{i=1}^n(1+a_i)\leq1+2\sum_{i=1}^na_i$

Question

Let $a_1,..,a_n$ be non negative real numbers , $a_i\geq0$ and $\sum _{i=1}^na_i\leq1$ Then $$\prod_{i=1}^n(1+a_i)\leq1+2\sum_{i=1}^na_i $$

I want to prove it by induction . For n=1 it is clear .

Now I have $$ \prod_{i=1}^{n+1}(1+a_i)\leq(1+a_{n+1})(1+2\sum_{i=1}^na_i)=1+2\sum_{i=1}^na_i+a_{n+1}+2a_{n+1}\sum_{i=1}^na_i=$$

$$=1+2\sum_{i=1}^{n+1}a_i-a_{n+1}+2a_{n+1}\sum_{i=1}^na_i $$

My problem is to show that $-a_{n+1}+2a_{n+1}\sum_{i=1}^na_i\leq0$ .

Thanks for the help .

Answer 1

Do you need to use induction? Alternatively, take the natural logarithm of both sides, so it is equivalent to showing \begin{equation} \sum_{i=1}^n \ln(1+a_i)\leq \ln(1+2\sum_{i=1}^n a_i). \end{equation} Use the following two inequalities: for $x\geq 0$, $\ln(1+x)\leq x$ and for $0\leq x\leq 2$, $\ln(1+x)\geq x/2$. You can check that both hold numerically.

Answer 2

We use Mathematical Induction to prove the stronger inequality $$\prod_{i=1}^n (1 + a_i) \le 1 + \left(1 + \sum_{i=1}^n a_i\right)\sum_{i=1}^n a_i.$$

For $n = 1$, it is true.

Assume that it is true for $n$ ($n\ge 1$).

For $n+1$, by the inductive hypothesis, we have \begin{align*} \prod_{i=1}^{n+1} (1 + a_i) &\le \left[1 + \left(1 + \sum_{i=1}^n a_i\right)\sum_{i=1}^n a_i\right](1 + a_{n+1}). \end{align*} It suffices to prove that $$ \left[1 + \left(1 + \sum_{i=1}^n a_i\right)\sum_{i=1}^n a_i\right](1 + a_{n+1}) \le 1 + \left(1 + \sum_{i=1}^{n+1} a_i\right)\sum_{i=1}^{n+1} a_i. \tag{1}$$ Let $S = \sum_{i=1}^n a_i, x = a_{n+1}$. We have $$\mathrm{RHS}_{(1)} - \mathrm{LHS}_{(1)} = Sx - S^2x + x^2 \ge 0.$$

We are done.