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Suppose $a_k > 0$ for $k=1,...,n$ and $\sum_{k=1}^n \dfrac{1}{1+a_k} = n-1 $. Prove that

$$ \prod_{k=1}^n \dfrac{1}{a_k} \geq (n-1)^n $$

Proof:

Taking $log$ in both sides we obtain that

$$ \sum_{k=1}^n \log(1/a_k) \geq n \log (n-1) $$

And by hypothesis, it be enough if we can prove

$$ \sum_{k=1}^n \log \left( \dfrac{1}{a_k } \right) \geq n \log \left( \sum_{k=1}^n \dfrac{1}{1+a_k} \right) $$

I dont see how to prove this. But, we go back to the beginning and apply Harmonic-Geometric mean ineq to $( \prod 1/a_k )^{1/n} $, we obtain

$$ \left( \prod_{k=1}^n \dfrac{1}{a_k} \right)^{1/n} \geq \dfrac{n}{\sum_{k=1}^n a_k } \geq \sum_{k=1}^n \dfrac{1}{1+a_k} = n -1 $$

and this would be the end but we still to prove that

$$ n \geq (a_1+.... + a_n) \left( \dfrac{1}{1+a_1} + ... + \dfrac{1}{1+a_n} \right) $$

but I am unable to prove this. Any assistance would be greatly appreciated.

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James
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  • what you are trying to prove $n \geq (a_1+.... + a_n) \left( \dfrac{1}{1+a_1} + ... + \dfrac{1}{1+a_n} \right)\ $is not true. I give you a contradiction. what you want to prove is same as proving $ \ \frac{n}{n-1}\geq (a_1+.... + a_n) $. Let us assume $\ n\geq2\ $. You can safely assume one of the variables, let's say $ a_1$ to be $n$ and rest other such that they follow $a_k>0$ and $\sum_{k=1}^n \dfrac{1}{1+a_k} = n-1$. But see here that the sum $(a_1+.... + a_n)\geq n\geq \dfrac{n}{n-1}$ – udit narayan pandey May 03 '20 at 05:46

1 Answers1

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Note that by Titu's lemma applied to the condition, we have $ \sum a_i \geq \frac{n}{n-1} $, so your HM-GM is too strong already.

The condition is equivalent to $\sum \frac{1}{ 1 + \frac{1}{a_i}} = 1$.

$ \sum \frac{1}{ 1 + 1/a_i } = \sum \frac{ a_i } { 1 + a_i } = \sum 1 - \frac{1}{1+a_i } = n - (n-1) = 1.$

Let $ b_i = \frac{1}{ 1 + \frac{1}{a_i}}$, so $ \frac{1}{a_i} = \frac{1-b_i } { b_i} $ and $ \sum b_i = 1$.

Thus $ \prod \frac{1}{a_i} = \prod \frac{ 1 - b_ i } { b_i } = \prod \frac{ \sum_{j\neq i } b_j } { b_i } \geq \prod \frac{ (n-1) \prod_{i\neq j } b_j ^ \frac{1}{n-1} } { b_i } = (n-1)^n $.

Several variants of this problem has appeared in various olympiads.

E.g. Under the same conditions, show that $ \sum \frac{1}{ \sqrt{a_i} } \geq (n-1) \sum \sqrt{a_i} $

Calvin Lin
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