Question

If a² + b² + c² - ab - bc - ca ≤ 0 then what is the value of (a +b)/c is .​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca \leqslant 0$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{a + b}{c}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca \leqslant 0$

Multiply by 2, on both sides, we get

$\rm :\longmapsto\: 2{a}^{2} + 2{b}^{2} + 2{c}^{2} - 2ab - 2bc - 2ca \leqslant 0$

can be rewritten as

$\rm :\longmapsto\: {a}^{2} + {a}^{2} + {b}^{2} + {b}^{2} + {c}^{2} + {c}^{2} - 2ab - 2bc - 2ca \leqslant 0$

can be re-arranged as

$\rm :\longmapsto\:( {a}^{2} + {b}^{2} - 2ab) + ( {b}^{2} + {c}^{2} - 2bc) + ( {c}^{2} + {a}^{2} - 2ac) \leqslant 0$

$\rm :\longmapsto\: {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} \leqslant 0$

We know, sum of squares can never be negative.

So

$\rm :\longmapsto\: {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0$

Now, we know that, sum of squares is zero only, when term itself is 0.

$\rm :\implies\:a - b = 0 \: \: and \: \: b - c = 0 \: \: and \: \: c - a = 0$

$\rm :\implies\:a = b \: \: and \: \: b = c \: \: and \: \: c = a$

$\bf\implies \:a = b = c$

So,

Let assume that

$\red{\rm :\longmapsto\:a = b = c = x}$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{a + b}{c}$

$\rm \: = \: \: \dfrac{x + x}{x}$

$\rm \: = \: \: \dfrac{2x}{x}$

$\rm \: = \: \: 2$

Hence,

$\bf :\longmapsto\:\dfrac{a + b}{c} = 2$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)