Question

If x, y, z are non negative real numbers such that x+y+z=1, then find the minimum value of
$( {x}^{ - 1} + 1)( {y}^{ - 1 } + 1)( {z}^{ - 1} + 1)$
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Answer 1

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

Given that,

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

and

$\rm :\longmapsto\:x + y + z = 1$

Now, we have to find minimum value of

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

can be rewritten as

$\rm \:  =  \:\bigg[\dfrac{1}{x} + 1 \bigg]\bigg[\dfrac{1}{y} + 1\bigg]\bigg[\dfrac{1}{z} +1 \bigg]$

$\rm \:  =  \:\bigg(\dfrac{1}{x} + 1 + \dfrac{1}{xy} + \dfrac{1}{y} \bigg)\bigg[\dfrac{1}{z} + 1\bigg]$

$\rm \:  =  \:\dfrac{1}{x} + 1 + \dfrac{1}{xy} + \dfrac{1}{y} + \dfrac{1}{xz} + \dfrac{1}{z} + \dfrac{1}{xyz} + \dfrac{1}{yz}$

can be re-arranged as

$\rm \:  =  \:1 + \dfrac{1}{xyz} + \bigg(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg) + \bigg(\dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{zx} \bigg)$

$\rm =1 + \dfrac{1}{xyz} + \bigg(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg) + \bigg(\dfrac{x + y + z}{xyz} \bigg)$

$\rm =1 + \dfrac{1}{xyz} + \bigg(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg) + \bigg(\dfrac{1}{xyz} \bigg)$

$\red{\bf =1 + \dfrac{2}{xyz} + \bigg(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg) - - - - (1)}$

Now, Consider,

$\rm :\longmapsto\:x + y + z = 1$

We know,

$\boxed{ \rm \:AM \geqslant GM}$

So, for the non zero real numbers, x, y, z, we have

$\rm :\longmapsto\:\dfrac{x + y + z}{3} \geqslant \sqrt[3]{xyz}$

$\rm :\longmapsto\:\dfrac{1}{3} \geqslant \sqrt[3]{xyz}$

On Cubing both sides,

$\rm :\longmapsto\:\dfrac{1}{27} \geqslant xyz$

$\bf\implies \:\dfrac{1}{xyz} \geqslant 27$

Now,

Also we know, that,

$\boxed{ \rm \:AM \geqslant HM}$

So, for the non zero real numbers,

$\rm :\longmapsto\:\dfrac{1}{x}, \dfrac{1}{y} ,\dfrac{1}{z} \: we \: have$

$\rm :\longmapsto\:\dfrac{\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} }{3} \geqslant \dfrac{3}{\dfrac{1}{\dfrac{1}{x} } + \dfrac{1}{\dfrac{1}{y} } + \dfrac{1}{\dfrac{1}{z} } }$

$\rm :\longmapsto\:\dfrac{\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} }{3} \geqslant \dfrac{3}{x + y + z}$

$\rm :\longmapsto\:\dfrac{\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} }{3} \geqslant 3$

$\bf\implies \:\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \geqslant 9$

So, substituting these values evaluated above in equation (1), we get

$\red{\bf =1 + \dfrac{2}{xyz} + \bigg(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg) - - - - (1)}$

$\rm \: \geqslant \:1 + 2(27) + 9$

$\rm \: \geqslant \:1 + 54 + 9$

$\rm \: \geqslant \:64$

So,

$\rm :\longmapsto\:( {x}^{ - 1} + 1)( {y}^{ - 1 } + 1)( {z}^{ - 1} + 1) \geqslant 64$

Hence,

$\boxed{ \rm \:\:Minimum \: value \: of \: ( {x}^{ - 1} + 1)( {y}^{ - 1 } + 1)( {z}^{ - 1} + 1) \: is \: 64}$