Let x<1and y<1
the sum of inequalities is also a true inequality that is
x+y<2
If (x<1)∧(y<1) then∑(x+y)<2
The contrapositive version:
if ∑(x+y)≥2; then¬((x<1)∧(y<1))
¬(x<1) is (x≥1); ¬(y<1) is (y≥1);
according to de Morgan’s laws:
¬((x<1)∧(y<1))=¬(x<1)∨¬(y<1)=(x≥1)∨(y≥1)
if ∑(x+y)⪖2 then(x≥) or (y≥1)
proof done
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