Olympiad Problem 2
x 3 ( y + z ) 1 + y 3 ( x + z ) 1 + z 3 ( x + y ) 1
Let x , y and z be positive reals such that x y z = 1 . If the minimum value of the expression above can be expressed in the form b a , find the value of a − b .
The answer is 1.
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2 solutions
Same thing but starting approach is slightly different -
x 2 ( x y + x z ) 1 + y 2 ( x y + y z ) 1 + z 2 ( x z + y z ) 1
x 2 . y 2 . z 2 ( x y + x z ) y 2 . z 2 + x 2 y 2 z 2 ( x y + y z ) x 2 . z 2 + x 2 . y 2 . z 2 ( x z + y z ) x 2 . y 2
I am amazed at your elegant proofs. I did the same way till the AM-GM part where I got bored and just plugged in equality case... :P Well, how do you know so much awesome proof-writing? @Sean Ty .... @Kartik Sharma -How did you do it? ...BTW-Just found out that this is an IMO problem! Whoa!!!!!!
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It's not that awesome, I have seen IMO problems that are way harder than this. Actually, I'm practicing by collecting some previous problems that appeared in the IMO and answer them. If you would ask me why this was an IMO problem, it's because Cauchy-Schwarz in Engel form didn't exist yet (that time). :)
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Hey! @Sean Ty - I have a crazy doubt- I will now demonstrate proof of a "relatively simple ineq.that uses AM-GM"..Uhave to say whether this kind of thing is accpetable- Like the question is P.T- If a,b,c (non negative real) and a+b+c=3, then a^2+b^+c^2+ab+bc+ac>=6. For this I first wrote the expression as (a+b+c)^2- (ab+Bc+ac)- Now this is equal to 9- (ab+bc+ac) Then I used some kind of logic plus ineuqality that ab+bc+ac>=a+b+C thus I substituted the minimum of this over there and finished the proof...SOrry if this was too silly/crazy :P
Actually you can just assume values of x,y,z as 1. After adding you will get 3/2. Therefore a-b=1
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That's the equality case for AM-GM, Cauchy-Schwarz, etc. But it does not work on all inequality problems.
Let: x 1 = a , y 1 = b , z 1 = c . . Also a b c = 1 and now applying Titu's Theorem:
M = b + c a 2 + a + c b 2 + a + b c 2 > = 2 ( a + b + c ) ( a + b + c ) 2
Applying AM-GM: M > = 2 a + b + c > = 2 3 3 a b c = 2 3
So: 3 − 2 = 1
We can simplify this equation into c y c ∑ x ( y + z ) x 2 1
By Cauchy-Schwarz in Engel Form, we have
c y c ∑ x ( y + z ) x 2 1 ≥ 2 ( x y + y z + z x ) ( x 1 + y 1 + z 1 ) 2
Upon simplification we get, 2 ( x y + y z + z x ) ( x 1 + y 1 + z 1 ) 2 = 2 x y + y z + z x
By AM-GM,
2 x y + y z + z x ≥ 2 3 3 x 2 y 2 z 2 = 2 3
Which can be achieved when x = y = z = 1
And so a − b = 1 .