Question

if x is real , then minimum value of x^2-x+1/x^2+x+1 is​

Answer 1

Value will be minimum when denominator is max ,
At max denominator , derivative of denominator will tend to zero
I.e , der(x^2+x+1) = 0
-> 2x+1=0
-> x=-1/2
So minimum at x=-1/2
Min=3.5

Answer 2

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

Let assume that

$\rm :\longmapsto\:y = \dfrac{ {x}^{2} - x + 1 }{ {x}^{2} + x + 1}$

$\rm :\longmapsto\:y( {x}^{2} + x + 1) = {x}^{2} - x + 1$

$\rm :\longmapsto\:y{x}^{2} + yx + y = {x}^{2} - x + 1$

$\rm :\longmapsto\:y{x}^{2} + yx + y - {x}^{2} + x - 1 = 0$

$\rm :\longmapsto\: {x}^{2}(y - 1) + x(y + 1) + y - 1 = 0$

Now, For x to be real,

$\rm :\longmapsto\:Discriminant,D \geqslant 0$

$\rm :\longmapsto\: {b}^{2} - 4ac = 0$

Here,

$\rm :\longmapsto\:a = y - 1$

$\rm :\longmapsto\:b = y + 1$

$\rm :\longmapsto\:c= y - 1$

So,

$\rm :\longmapsto\: {(y + 1)}^{2} - 4(y - 1)(y - 1) \geqslant 0$

$\rm :\longmapsto\: {(y + 1)}^{2} - {[2(y - 1)]}^{2} \geqslant 0$

$\rm :\longmapsto\: {(y + 1)}^{2} - {[2y - 2]}^{2} \geqslant 0$

$\rm :\longmapsto\:(y + 1 + 2y - 2)(y + 1 - 2y + 2) \geqslant 0$

$\rm :\longmapsto\:(3y - 1)(3 - y) \geqslant 0$

$\rm :\longmapsto\: - (3y - 1)( y - 3) \geqslant 0$

$\rm :\longmapsto\: (3y - 1)( y - 3) \leqslant 0$

$\bf\implies \:\dfrac{1}{3} \leqslant y \leqslant 3$

$\bf\implies \:\dfrac{1}{3} \leqslant \dfrac{ {x}^{2} - x + 1 }{ {x}^{2} + x + 1} \leqslant 3$

Hence,

$\bf\implies \:Minimum \: value \: of \: \dfrac{ {x}^{2} - x + 1 }{ {x}^{2} + x + 1} = \dfrac{1}{3}$

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MORE TO KNOW

Let a and b are two real numbers such that a < b then

$\boxed{\tt{ (x - a)(x - b) < 0 \: \rm\implies \:a < x < b}}$

$\boxed{\tt{ (x - a)(x - b) \leqslant 0 \: \rm\implies \:a \leqslant x \leqslant b}}$

$\boxed{\tt{ (x - a)(x - b) > 0 \: \rm\implies \:x < a \: or \:x > b}}$

$\boxed{\tt{ (x - a)(x - b) \geqslant 0 \: \rm\implies \:x \leqslant a \: or \:x \geqslant b}}$