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Consider set $S = \left \{ a_i \mid 1 \leq i \leq n \right \}$ where $a_i \in \mathbb{R}$ and fixed integer $t$

I do not know whether the following inequality is true or not:

$$\sum_{P \subset S, \left | P \right | = t}\left ( {\prod_{j=1}^{\left | P \right |} a_j}\right ) \leqslant \left ( \sum_{i=1}^{\left | S \right |} \frac{a_i}{n}\right )^{t} \binom{n}{t}$$

i.e. approximating each $a_i$ by the average $ \sum_{i=1}^{n} {\frac{a_i}{n}}$ and with $\binom{n}{t}$ number of such sets P

For example: $S = \left \{ a,b,c \right \}$ and $t=2$, The above inequality states :

$ab+bc+ca \leqslant {3\left ( \frac{a+b+c}{3} \right )}^{2}$

I am able to conclude the validity for this example as follows:

$2(ab+bc+ca) = (a+b+c)^2 - (a^2 + b^2 + c^2)$

Now using Cauchy Schwarz's Inequality, we have $ (a+b+c)^2\leqslant 3(a^2 + b^2 + c^2)$

$2(ab+bc+ca) \leqslant (a+b+c)^2 - \frac{(a+b+c)^2}{3}$

$\therefore ab+bc+ca \leqslant {3\left ( \frac{a+b+c}{3} \right )}^{2}$

Please, tell if this inequality is true for general case or not? Thank you. and Sorry for the bad title for the question (Please suggest some).

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    $\begingroup$ The inequaity $$ab+bc+ca \leqslant {3\left ( \frac{a+b+c}{3} \right )}^{2}$$ is just Chebyshev ;) $\endgroup$ – N. S. Aug 12 '18 at 17:31
  • $\begingroup$ I think you mean "Chebyshev's sum inequality" ? Thanks, I learned new inequality today :) $\endgroup$ – Inuyasha yagami Aug 12 '18 at 18:10
  • $\begingroup$ But, again this inequality holds for a,b,c,d and t = 2 also, which we may not get using Chebyshev's. $\endgroup$ – Inuyasha yagami Aug 12 '18 at 18:11
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If all $a_i$ are assumed to be non-negative then it is an immediate consequence of Maclaurin's inequality. In fact,

$$ S_t = \frac{\sum_\limits{P \subset S, | P | = t}\left ( {\prod_\limits{a \in P} a}\right )}{ \binom{n}{t}} \, , $$ satisfies $$ \sqrt[t]{S_t} \le S_1 = \frac 1n \sum_{i=1}^{n} a_i \, . $$

Without that assumption it does not hold in general. A counter-example is $S = \{ 2, -1, -1 \}$ and $t = 3$.

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