What Is the Minimum Sum of $a+b+c+d+e+f+g$ Given Their Progressive Inequalities?

[anemone]

Here is this week's POTW:

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Find the smallest possible value of $a+b+c+d+e+f+g$ if $a,\,b,\,c,\,d,,\,e,\,f$ and $g$ are positive integers that satisfy

$a<b<c<d<e<f<g<a^2<b^2<c^2<d^2<e^2<f^2<g^2<a^3<b^3<c^3<d^3<e^3<f^3<g^3$

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[anemone]

Congratulations to the following members for their correct solution::)

1. castor28
2. kaliprasad
3. lfdahl

Solution from castor28:

Spoiler

To get the smallest sum compatible with the constraints, we should take $a,\ldots,g$ as consecutive integers. Indeed, with another choice, we could close the gaps and get a smaller sum while still satisfying the constraints.

With that choice, the constraints $a^n < \ldots < g^n$ will be automatically satisfied, since $x^n$ is an increasing function of $x$ for $n>0$. We must still satisfy the constraints:

$a^2 > g = a+6$
$a^3 > g^2 = (a+6)^2$
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The first relation gives $a>3$, and the second one gives $a>4.923$. We should therefore take $a=5, b = 6, \ldots, g=11$, and the sum is equal to $\dfrac{7(5+11)}{2} = 56$.