If a^{2}+b^{2}+4c^{2} = ab+2bc+2ca. Prove that a = b = 2c

2. a^{2} + 2. b^{2} + 2.4c^{2} = 2.ab + 2.2bc + 2.2ca

2. a^{2} + 2. b^{2} + 2.4c^{2} = 2ab + 4bc + 4ca

2. a^{2} + 2. b^{2} + 2.4c^{2} - 2ab - 4bc - 4ca =0

( a^{2} - 2ab + b^{2} ) + ( b^{2} - 4bc + 4c^{2} ) + ( a^{2} - 4ca + 4c^{2} ) = 0

( a - b )^{2} + (b - c)^{2} + (a- 2c)^{2} = 0

If the Sum of the squares will be equal to zero then the individual terms will also be equal to zero.

∴ ( a - b )^{2} = 0

(b - c)^{2} = 0

and (a- 2c)^{2} = 0

Now ( a - b )^{2} = 0 , or ( a - b ) = 0 or a = b

Similarly (b - c)^{2} = 0 , or (b - c) = 0 or b = c

Similarly (a- 2c)^{2} = 0 , or (a- 2c) = 0 . or a = 2c

∴ a = b = 2c __Proved__

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