Let $a_1, a_2,\ldots, a_{100}$ be non-zero real numbers such that$a_1+ a_2+\cdots+ a_{100}=0$. Then $\sum_{i=1}^{100} a_i 2^{a_i}$

Question

Let $a_1, a_2,\ldots, a_{100}$ be non-zero real numbers such that $$a_1 + a_2 + \cdots + a_{100} = 0$$ Then

A - $\sum_{i=1}^{100} a_i 2^{a_i}\ge0$ and $\sum_{i=1}^{100} a_i 2^{-a_i}\ge0$

B- $\sum_{i=1}^{100} a_i 2^{a_i}\gt0$ and $\sum_{i=1}^{100} a_i 2^{-a_i}\lt0$

C $\sum_{i=1}^{100} a_i 2^{a_i}\le0$ and $\sum_{i=1}^{100} a_i 2^{-a_i} \le 0$

D The sign depend upon the choice of $a_i$

I tried adding $\sum_{i=1}^{100} a_i 2^{a_i}$ and $\sum_{i=1}^{100} a_i 2^{-a_i}$ but I am stuck. No idea what to do next.

P.S: The option B was earlier typed incorrectly. Corrected the typing mistake.

Answer 1

I think the following reasoning solves all problems. $$\sum_{i=1}^{100}a_i2^{a_i}=\sum_{i=1}^{100}\left(a_i2^{a_i}-a_i\right)=\sum_{i=1}^{100}a_i\left(2^{a_i}-1\right)\geq0$$ because for all real $x$ we have $x(2^x-1)\geq0.$

The equality occurs for $a_1=a_2=...=a_{100}=0$, which says that the equality does not occur.

Id est, $$\sum_{i=1}^{100}a_i2^{a_i}>0.$$

By the same way we can get that $$\sum_{i=1}^{100}a_i2^{-a_i}<0.$$

Answer 2

Hint: Well, if $a_i$ satisfy the condition, then so does $-a_i$. Then note that multiplying by $-1$ should reverse the inequality, whichever it is!

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P.S. adding more detail

If possible, say $\sum a_i 2^{a_i} \geqslant 0$ for all $a_i$ satisfying the condition. Then $\{-a_i\}$ also satisfy the same condition, so we must have $\sum -a_i2^{-a_i} \geqslant 0$. Multiplying by $-1$, we get $\sum a_i 2^{-a_i} \leqslant 0$. The same argument holds if either $A, B, C$ are taken as hypothesis, so none of them can be true.