Let x and y be real numbers. Prove that, if 5x+y>11, then x>2 or y>1.
for x = y:
6y>11 ⟹ y>11/6 ⟹ y>16y>11\implies y>11/6 \implies y>16y>11⟹y>11/6⟹y>1
for x > y:
x=y+k,k>0x=y+k,k>0x=y+k,k>0
6x−k>11 ⟹ x>(11+k)/66x-k>11\implies x>(11+k)/66x−k>11⟹x>(11+k)/6
6y+5k>116y+5k>116y+5k>11
if y≤1y\le 1y≤1 then k>1 ⟹ x>2k>1\implies x>2k>1⟹x>2
for x < y:
x=y−k,k>0x=y-k,k>0x=y−k,k>0
6x+k>116x+k>116x+k>11
6y−5k>11 ⟹ y>11/6+5k/6>16y-5k>11\implies y>11/6+5k/6>16y−5k>11⟹y>11/6+5k/6>1
so, if 5x+y>11, then x>2 or y>1
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