Question

If x,y,z are real numbers such that x + y + z = 4 and x^2 + y^2 + z^2 = 6, prove that each x, y, z lies in closed interval [2/3, 2]

Please don't spam.

Need quality answer.

Please don't answer if you don't know.

Don't greedy for 5 marks.

Let me learn.​

Answer 1

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

Given that,

$\rm :\longmapsto\:x + y + z = 4 - - - (1)$

and

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + {z}^{2} = 6 - - - - (2)$

From equation (1), we get

$\rm :\longmapsto\:z = 4 - x - y - - - - (3)$

On substituting the value of z, in equation (2), we get

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + {(4 - x - y)}^{2} = 6$

$\rm :\longmapsto\: {x}^{2}+ {y}^{2} +16+{x}^{2}+{y}^{2} - 8x + 2xy - 8y = 6$

$\rm :\longmapsto\: 2{x}^{2}+ 2{y}^{2} +10 - 8x + 2xy - 8y = 0$

$\rm :\longmapsto\: {x}^{2}+ {y}^{2} +5 - 4x + xy - 4y = 0$

$\rm :\longmapsto\: {x}^{2} + x(y - 4)+ {y}^{2}- 4y + 5 = 0$

For x to be real, the Discriminant of quadratic

$\red{ \boxed{ \sf{ \:Discriminant, \: {b}^{2} - 4ac \geqslant 0}}}$

$\rm :\longmapsto\: {(y - 4)}^{2} - 4( {y}^{2} - 4y + 5) \geqslant 0$

$\rm :\longmapsto\: {y}^{2} + 16 - 8y - 4{y}^{2} + 16y - 20 \geqslant 0$

$\rm :\longmapsto\:- 3{y}^{2} + 8y - 4 \geqslant 0$

$\rm :\longmapsto\:- (3{y}^{2} - 8y + 4) \geqslant 0$

$\rm :\longmapsto\:3{y}^{2} - 8y + 4 \leqslant 0$

$\rm :\longmapsto\:3{y}^{2} - 6y - 2y + 4 \leqslant 0$

$\rm :\longmapsto\:3y(y - 2) - 2(y - 2) \leqslant 0$

$\rm :\longmapsto\:(y - 2)(3y - 2) \leqslant 0$

$\bf\implies \:\dfrac{2}{3} \leqslant y \leqslant 2$

$\bf\implies \:y \: \in \: \bigg[\dfrac{2}{3}, \: 2 \bigg]$

Hence, Proved

Now, from equation (1),

$\rm :\longmapsto\:y = 4 - x - z - - - - (4)$

On substituting equation (4) in equation (2), we get

$\rm :\longmapsto\: {x}^{2} + {z}^{2} + {(4 - x - z)}^{2} = 6$

$\rm :\longmapsto\: {x}^{2}+ {z}^{2} +16+{x}^{2}+{z}^{2} - 8x + 2xz - 8z = 6$

$\rm :\longmapsto\: 2{x}^{2}+ 2{z}^{2} +10 - 8x + 2xz - 8z = 0$

$\rm :\longmapsto\: {x}^{2}+ {z}^{2} +5 - 4x + xz - 4z = 0$

$\rm :\longmapsto\: {x}^{2} + x(z - 4)+ {z}^{2}- 4z + 5 = 0$

For real values of x, Discriminant of quadratic equation

$\red{ \boxed{ \sf{ \:Discriminant, \: {b}^{2} - 4ac \geqslant 0}}}$

$\rm :\longmapsto\: {(z - 4)}^{2} - 4( {z}^{2} - 4z + 5) \geqslant 0$

$\rm :\longmapsto\: {z}^{2} + 16 - 8z - 4{z}^{2} + 16z - 20 \geqslant 0$

$\rm :\longmapsto\:- 3{z}^{2} + 8z - 4 \geqslant 0$

$\rm :\longmapsto\:- (3{z}^{2} - 8z + 4) \geqslant 0$

$\rm :\longmapsto\:3{z}^{2} - 8z + 4 \leqslant 0$

$\rm :\longmapsto\:3{z}^{2} - 6z - 2z + 4 \leqslant 0$

$\rm :\longmapsto\:3z(z - 2) - 2(z - 2) \leqslant 0$

$\rm :\longmapsto\:(z - 2)(3z - 2) \leqslant 0$

$\bf\implies \:\dfrac{2}{3} \leqslant z \leqslant 2$

$\bf\implies \:z \: \in \: \bigg[\dfrac{2}{3}, \: 2 \bigg]$

Similarly,

$\bf\implies \:y \: \in \: \bigg[\dfrac{2}{3}, \: 2 \bigg]$