Question

plz solve it with steps
BEST ANSWER WILL BE MARKED AS BRAINLIEST

Answer 1

since,

a² + b² + c² - ab - bc - ca

let us multiply the equation by 2,

then,

2 × ( a² + b² + c² - ab - bc - ca ) ÷ 2

=>(2a² + 2b² + 2c² - 2ab - 2bc - 2ca ) ÷ 2

it can be written as,

=> (a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca ) ÷ 2

so, to form an identity, we will group them,

=> {(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) } ÷ 2

then, using the identity,

(a - b)² = a² - 2ab + b²

we will solve this equation,

=>{ ( a - b )² + ( b - c )² + ( c - a )² } ÷ 2

so, when we do square of any number , it's result is always positive!

and, the sum of these terms will be also positive because sum of positive numbers is also positive!

therefore,

For any value of a,b,c 

(a-b)² ≥ 0,

(b-c)² ≥ 0,

(c-a)² ≥ 0,

So, 

a²+b²+c²-ab-bc-ca = ½ { (a-b)² + (b-c)² + (c-a)² } ≥ 0, i.e. Non-negative [ proved ]

Answer 2

this is not answer of this question