Question

If a² + b² + c² - ab – bc - ca <= 0, (where a, b, c are non-zero real number) then value of
is : (a+b)/c​

Answer 1

Answer:

2

Step-by-step explanation:

given:

a² + b² + c² - ab – bc - ca <= 0

multiply with 2 on both sides,

2a²+2b²+2c² -2ab-2bc-2ca ≤ 0

a²+b²-2ab+ b²+c²-2bc + c²+a²- 2ac ≤ 0

(a-b)² + (b-c)² + (c-a)² ≤0

all (a-b)², (b-c)², (c-a)² are ≥0 . so above equation cannot be negative. so given equation should be equal to 0.

and as there are no negative parts, each of (a-b)², (b-c)², (c-a)² should be equal to 0.

a-b =0, b-c =0, c-a = 0

so, a=b=c

now, (a+b)/c = 2a/a

= 2

I tried my level best to explain.

Answer 2

2

Explanation:

Given a² + b² + c² - ab – bc - ca <= 0

Multiply 2 on both sides

          2a² + 2b² + 2c² - 2ab – 2bc - 2ca <= 0

(a² + b² - 2ab) + (b² + c² - 2bc) + (a² + c² - 2ca) <= 0

(a-b)² + (b-c)² + (c-a)² ≤0

Square of any real number can not be negative.

Therefore,  (a-b)² + (b-c)² + (c-a)² = 0

        a-b = 0, a = b

        b-c = 0, b = c

        c-a = 0, c = a

That implies, a=b=c

Value of  (a+b)/c​  =  (a+a)/a​

                            =  (2a)/a

Value of  (a+b)/c = 2

#SPJ2

https://brainly.in/question/40626097

https://brainly.in/question/42408558