Question

Pay attention !!

$if \:$
x , y , z are all positive real numbers . prove that

$(x + y + z) (1 \div x + 1 \div y + 1 \div z) \geqslant 9$
( IIT 1984 )

Answer 1

$\bf HELLO \: \: DEAR,$

$\bf (x + y + z)(1/x + 1/y + 1/z) \geqslant 9 \\ \\ \bf (x + y + z)(xy + yz + xz)/xyz - 9 \geqslant 0 \\ \\ \bf [(x + y + z)(xy + yz + xz) - 9xyz] / xyz \geqslant 0 \\ \\ \bf [ x^{2}y + x^{2}z + xy^{2} - 6xyz + xz^{2} + y^{2}z + yz^{2}]/xyz \geqslant 0 \\ \\ \bf HENCE,\:\: the \:\:smallest\:\: value \:\:for \:\:x , \:\:y ,\:\: z \:\:is (1)$

$\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$

Answer 2

Hey Ranjan Bhaiya !

Here is your solution :

=> ( x + y + z ) [ ( 1/x ) + ( 1/y ) + ( 1/z ) ] ≥ 9

=> { x ( 1/x ) } + { x ( 1/y ) } + { x ( 1/z ) } + { y ( 1/x ) } + { y ( 1/y ) } + { y ( 1/z ) } + { z ( 1/x ) } + { z ( 1/y ) } + { z ( 1/z ) } ≥ 9

=> 1 + ( x/y ) + ( x/z ) + ( y/x ) + 1 + ( y/z ) + ( z/x ) + ( z/y ) + 1 ≥9

Rearranging the terms ,

=> 3 + ( x/y ) + ( y/x ) + ( x/z ) + ( z/x ) + ( y/z ) + ( z/y ) ≥ 9

Now,

To prove this , we need to prove that ( x/y ) + ( y/x ) ≥ 2 and so on.

=> ( x/y ) + ( y/x ) ≥ 2

=> ( x² + y² )/ xy ≥2

=> ( x² + y² ) / xy ≥ 2

=> ( x² + y² ) ≥ 2xy

=> x² + y² - 2xy ≥ 0

=> ( x - y )² ≥ 0

We know that square of difference of any 2 numbers is always greater than or equal to 0.

So, ( x/y ) + ( y/x ) ≥ 2.

In same way,

( x/z ) + ( z/x ) ≥ 2

And ,

( y/z ) + ( z/y ) ≥ 2.

By substituting their value ,

= 3 + ( ≥2 ) + ( ≥2 ) + ( ≥2 )

=> ≥ 9 = L.H.S

Proved.
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Hope it helps !!