Finding the minimum sum

All numbers considered are positive real.


M spaces exist

In each space, N sets of the form (a,b,c) exist.

Now we have the following equations: \[({a_{11}} + {b_{11}} + {c_{11}}) + ({a_{21}} + {b_{21}} + {c_{21}}) + ... + ({a_{N1}} + {b_{N1}} + {c_{N1}}) = {S_1}\] for space 1 … … (a1M+b1M+c1M)+(a2M+b2M+c2M)+...+(aNM+bNM+cNM)=SM({a_{1M}} + {b_{1M}} + {c_{1M}}) + ({a_{2M}} + {b_{2M}} + {c_{2M}}) + ... + ({a_{NM}} + {b_{NM}} + {c_{NM}}) = {S_M} for space M

We know that the sum SYS_Y is minimum. Then can we say that if we deduct a constant x from each middle element of each set of each space, then for the same combination of sets, we will get the minimum value? I.e. the new SYS_{Y} will remain the minimum sum?

The new equations after deduction are: (a11+b11x+c11)+(a21+b21x+c21)+...+(aN1+bN1x+cN1)=S1({a_{11}} + {b_{11}} - x + {c_{11}}) + ({a_{21}} + {b_{21}} - x + {c_{21}}) + ... + ({a_{N1}} + {b_{N1}} - x + {c_{N1}}) = {S_1} … … (a1M+b1Mx+c1M)+(a2M+b2Mx+c2M)+...+(aNM+bNMx+cNM)=SM({a_{1M}} + {b_{1M}} - x + {c_{1M}}) + ({a_{2M}} + {b_{2M}} - x + {c_{2M}}) + ... + ({a_{NM}} + {b_{NM}} - x + {c_{NM}}) = {S_M}

Note by Santanu Banerjee
5 months ago

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I suppose the answer is obviously yes. Remove the xx from the brackets, and you would get nxnx. Since SYSrandomS_Y\ge S_{\text{random}}, SYnxSrandomnxS_Y-nx\ge S_{\text{random}}-nx wouldn’t it? :)

Jeff Giff - 5 months ago

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