prove that a+b+2c a b
c b+c+2a b = 2( a+b+c)3
c a c+a+2b

Consider the following determinant.      a+b+2cabcb+c+2abcac+a+2bApply the column operation, C1 C1+ C2+ C3 to get,     =2a+b+cab2a+b+c  b+c+2ab2a+b+ca  c+a+2bTake 2a+b+c out from C1    =2a+b+c 1ab1  b+c+2ab1a  c+a+2bApply the row operations, R2R2-R1, R3R3-R1 to get,   =2a+b+c 1ab0  a+b+c000  a+b+cTake out a+b+c  from R2  and R3      =2a+b+ca+b+ca+b+c 1ab0  1000  1     =2a+b+c3  1ab0  1000  1Expand along C1 to get,    =2a+b+c3   1×1×1-0×0    =2a+b+c3   1-0    =2a+b+c3Thus, it is proved that,      a+b+2cabcb+c+2abcac+a+2b= 2a+b+c3      

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