Statement regarding primes $ \le n$ Following is the statement I believe is true, but can't prove.

Let $n$ be a natural. Let the primes less than equal to $\sqrt{n}$ be $p_1,p_2,...,p_k$. Let $\alpha_i$ be the greatest natural number such that $p_i^{\alpha_i} \le n$. Out of these $k$ primes let us take $2 \le r \le k$ primes namely $p_{j_1},p_{j_2},...,p_{j_r}$. Then there exist no natural numbers $b_1,b_2,..,b_r$ such that
  $$ \sum_{i=1}^{i=r}p_{j_i}^{\alpha_{j_i}} <  \prod_{i=1}^{i=r}p_{j_i}^{b_i} \le n  ...(1)$$

Example. Let $n=10$ then $k=2$, ie. we have $2$ primes less than $\sqrt{10}$ namely $2,3$ with the values of $\alpha$ being $3,2$ respectively. Now let us take $r=2$ primes, say $2$ and $3$, ie. take both the primes. The summation part of equation $(1)$ becomes $2^3+3^2=17$. The product in equation $(1)$ can only be $6$ in this case and hence clearly $6 < 17$. Thus there is no values of $b$ for which $(1)$ is true in this case.
PS: This question is in relation to this problem max-sum co-prime set. I would appreciate only hints.

Let $n = 124$, $p_1 = 2$, $p_2 = 5$. Then $\alpha_1 = 6$, since
$$2^6 = 64 \leq 124 < 2^7 = 128$$
and $\alpha_2 = 2$ since
$$5^2 = 25 \leq 124 < 5^3 = 125.$$
Now, $2^6 + 5^2 = 89$. Choose $b_1 = 2$ and $b_2 = 2$. Then
$$89 < 2^2\cdot 5^2 = 100 \leq 124 = n.$$
The answer below was for an earlier version of the question, before it was edited.
Let $n = 21$, take the primes $p_1 = 3$ and $p_2 = 7$. Then $\alpha_1 = 2$ since $3^2 = 9 \leq 21 < 3^3 = 27$ and $\alpha_2 = 1$ since $7 \leq 21 < 7^2 = 49$. Take $b_1 = b_2 = 1$. Then
$$9 + 7 = 16 < 3 \cdot 7 = 21 \leq 21 = n$$
