1.

show that the numbers 2b+2, b to the power 2 +2b, b to the power2 +2b +2 can represent sides of a right angled triangle​

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a2 + b2 + C2 − ab − bc − ca

Now we have to MULTIPLYING and dividing the expression by 2,

= 2(a2 + b2 + c2 − ab − bc – ca) / 2

= (2a2 + 2b2 + 2c2 − 2ab − 2bc − 2ca) / 2

= (a2 − 2ab + b2 + b2 − 2bc + c2 + c2− 2ca + a2) / 2

= [(a − b)2 + (b − c)2 + (c −  a)2] / 2

{ NOTE = Square of a number is always greater than or equal to zero.}

Hence SUM of the squares is also greater than or equal to zero

∴ [(a − b)2 + (b − c)2 + (c − a)2] ≥ 0  and {(a − b)2 + (b − c)2 + (c − a)2}/2= 0 when a = b = c.

NOW  a2 + b2 + c2 – ab – ac - bc is always non-negative for all its values of a, b and c.


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