# Inequality about sum of finitely many real numbers

Suppose that $a_1, a_2, ..., a_n$ are non-negative real numbers. Put $S = a_1 + a_2 + ... + a_n$. If $S < 1$, show that $$1+S\leq(1+a_1)...(1+a_n)\leq\dfrac{1}{1-S}.$$

I tried induction on $n$ and proved the left inequality but I can't prove the right inequality. It boils down to proving that $$(1+a_1)...(1+a_n)(1+a_{n+1})\leq\dfrac{1}{1-S-a_{n+1}}$$ where $S=a_1+...+a_n$. What should I do?

• for the right hand side, use this $$\frac{1}{1-S}=1+S+S^2+S^3+\dots$$ – Mher Nov 24 '14 at 14:12

$$(1+a_1)(1+a_2)\dots(1+a_n) =$$ $$=1 + (a_1+a_2+\dots+a_n) + (a_1a_2+a_1a_3+\dots+a_{n-1}a_n)+\dots+ (a_1a_2\dots a_n)\le$$ $$\le 1 + S + S^2 + \dots + S^n \le 1 + S + S^2 + \dots = \frac{1}{1-S}$$
Use $e^{a_i} \ge 1 + a_i$
and $1 \ge (1-x)e^x$ for $x \in [0,1]$