0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ for the right hand side, use this $$\frac{1}{1-S}=1+S+S^2+S^3+\dots$$ $\endgroup$ – Mher Nov 24 '14 at 14:12
0
$\begingroup$

Hint for the rigth side inequality:

$$(1+a_1)(1+a_2)\dots(1+a_n) =$$ $$=1 + (a_1+a_2+\dots+a_n) + (a_1a_2+a_1a_3+\dots+a_{n-1}a_n)+\dots+ (a_1a_2\dots a_n)\le$$ $$\le 1 + S + S^2 + \dots + S^n \le 1 + S + S^2 + \dots = \frac{1}{1-S}$$

$\endgroup$
  • $\begingroup$ So there is no need for induction, right? $\endgroup$ – oyster Nov 24 '14 at 14:46
  • $\begingroup$ Yeap! And in the same way left hand side inequality also can be done without induction. $\endgroup$ – Mher Nov 24 '14 at 14:54
0
$\begingroup$

Alternatively,

Use $e^{a_i} \ge 1 + a_i$

and $1 \ge (1-x)e^x$ for $x \in [0,1]$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.