Proposal for IMO

Algebra Level 4

Let x , y , z x, y, z be the real numbers in the interval [ − 1 , 2 ] \left[ -1, 2 \right] such that x + y + z = 0 x + y + z = 0 . Minimize

( 2 − x ) ( 2 − y ) ( 2 + x ) ( 2 + y ) + ( 2 − y ) ( 2 − z ) ( 2 + y ) ( 2 + z ) + ( 2 − z ) ( 2 − x ) ( 2 + z ) ( 2 + x ) . \large{ \sqrt { \frac {( 2 - x) ( 2 - y)}{( 2 + x)( 2 + y)}} + \sqrt { \frac {( 2 - y)( 2 - z ) }{( 2 + y) ( 2 + z)}} + \sqrt { \frac {( 2 - z ) ( 2 - x) }{( 2 + z )( 2 + x)}}}.


The answer is 3.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

If 0ne of them is 2, two terms will cansale out. Also note the limits [-1,2].

WLOG say z=2.

Again WLOG say x =0, y=-2, denominator becomes 0, rejected.

So let us take x=.000001, y=-1.999999. But y can only be -1.

So to be within given range x=y=-1.

So any one= 2, and remaining equal to -1,

E x p r e s i o n m i n = 3. Expresion_{min} =\Large \color{#D61F06}{3}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

They are real, so your proof is quite shoddy, for you to assume z=2. Therefore, try to generalise your solution

Alex Fullbuster - 2 years ago

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