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To prove: $(1+a_1)(1+a_2)\ldots(1+a_n)\geq\dfrac{2^n}{n+1}(1+a_1+a_2+\ldots+a_n)$ when $a_i\geq1$

This seems to be based on Bernoulli's Inequality (which can be proved by induction).

Trying the induction way, the given problem holds for $n=1$.

For $n=2$, I get $3a_1a_2\geq1+a_1+a_2$ which is true when $a_i\geq1$.

For $n=3$, I get $a_1a_2+a_2a_3+a_1a_3+a_1a_2a_3\geq1+a_1+a_2+a_3$ which is true.

But how do I take it up to general case ?

For every step, the RHS increases exponentially.

LHS grows like $1+\sum{a_i}+\sum\limits_{i\neq j}{a_ia_j}+\sum\limits_{i\neq j\neq k}{a_ia_ja_k}+\cdots$ which is polynomial.

So, does that make the problem wrong or is there some other way to look at this?

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let $a_{i}-1=b_{i},i=1,2,\cdots,n$, $$\Longleftrightarrow (b_{1}+2)(b_{2}+2)\cdots(b_{n}+2)\ge\dfrac{2^n}{n+1}(b_{1}+b_{2}+\cdots+b_{n}+n+1)$$ $$\Longleftrightarrow 2^n\left(1+\dfrac{b_{1}}{2}\right)\left(1+\dfrac{b_{2}}{2}\right)\cdots\left(1+\dfrac{b_{n}}{2}\right)\ge\dfrac{2^n}{n+1}(b_{1}+b_{2}+\cdots+b_{n}+n+1)$$ $$\Longleftrightarrow \left(1+\dfrac{b_{1}}{2}\right)\left(1+\dfrac{b_{2}}{2}\right)\cdots\left(1+\dfrac{b_{n}}{2}\right)\ge\dfrac{1}{n+1}(b_{1}+b_{2}+\cdots+b_{n})+1$$ use Bernoulli's inequality,we have $$(1+a_{1})(1+a_{2})\cdots(1+a_{n})\ge 1+a_{1}+a_{2}+\cdots+a_{n}$$ so $$\left(1+\dfrac{b_{1}}{2}\right)\left(1+\dfrac{b_{2}}{2}\right)\cdots\left(1+\dfrac{b_{n}}{2}\right)\ge 1+\dfrac{1}{2}(b_{1}+b_{2}+\cdots+b_{n})$$ and it is clear $$\dfrac{1}{2}(b_{1}+b_{2}+\cdots+b_{n})\ge\dfrac{1}{n+1}(b_{1}+b_{2}+\cdots+b_{n}),n\ge 1$$ By done!

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  • $\begingroup$ So beautiful .. Thanks a lot !! :) $\endgroup$ – square_one Jan 21 '15 at 6:01

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