Question

If x, y, z are real and distinct and positive, show that

${x}^{2} + {y}^{2} + {z}^{2} > xy + yz + zx$
​

Answer 1

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

Given that,

$\rm :\longmapsto\:x \: \ne \: y \: \ne \: z \: > \: 0$

We know,

Arithmetic mean between two distinct positive numbers is greater than Geometric mean.

Arithmetic mean between two positive numbers a and b is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ Arithmetic \: mean = \frac{a + b}{2} \: }}}$

Geometric mean between two positive numbers a and b is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ Geometric \: mean \: = \: \sqrt{ab} \: }}}$

So,

$\rm :\longmapsto\:Let \: consider \: two \: numbers \: {x}^{2} \: and \: {y}^{2}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + {y}^{2} }{2} > \sqrt{ {x}^{2} {y}^{2} }$

$\bf\implies \: {x}^{2} + {y}^{2} > 2xy - - - (1)$

Now,

$\rm :\longmapsto\:Let \: consider \: two \: numbers \: {y}^{2} \: and \: {z}^{2}$

$\rm :\longmapsto\:\dfrac{ {y}^{2} + {z}^{2} }{2} > \sqrt{ {y}^{2} {z}^{2} }$

$\bf\implies \: {y}^{2} + {z}^{2} > 2yz - - - - (2)$

Now,

$\rm :\longmapsto\:Let \: consider \: two \: numbers \: {x}^{2} \: and \: {z}^{2}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + {z}^{2} }{2} > \sqrt{ {x}^{2} {z}^{2} }$

$\bf\implies \: {x}^{2} + {z}^{2} > 2xz - - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm :\longmapsto\:2( {x}^{2} + {y}^{2} + {z}^{2}) > 2(xy + yz + zx)$

$\bf\implies \: {x}^{2} + {y}^{2} + {z}^{2} > xy + yz + zx$

Hence, Proved

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Additional Information

$\rm :\longmapsto\:Arithmetic \: mean \geqslant Geometric \: mean \geqslant Harmonic \: mean$

$\rm :\longmapsto\: {(Geometric \: mean)}^{2} = Arithmetic \: mean \times Harmonic \: mean$