Minimum value, you say?
y 3 + z 3 x 6 + z 3 + x 3 y 6 + x 3 + y 3 z 6
Let x , y and z be three positive real numbers satisfying x y z = 2 . Find the minimum value of the expression above.
The answer is 3.
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1 solution
is there a way to do this problem without titu's lemma? if not, can you explain to me what titu's lemma is? thanks!!!
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The only alternative solution I know is using calculus. Otherwise, I think there's none.
For reference, this is the wiki page about Titu's Lemma . :)
I had a confusion in AM-GM inequality(how did you dealed with the denominator??) . And in titu's lemma what did you did after we get 2 x 3 + y 3 + z 3 ???
Titu's Lemma gives y 3 + z 3 x 6 + z 3 + x 3 y 6 + x 3 + y 3 z 6 ≥ 2 ( x 3 + y 3 + z 3 ) ( x 3 + y 3 + z 3 ) 2 = 2 x 3 + y 3 + z 3 and by the AM-GM inequality, y 3 + z 3 x 6 + z 3 + x 3 y 6 + x 3 + y 3 z 6 ≥ 2 3 3 x 3 y 3 z 3 = 2 3 x y z = 2 3 ( 2 ) = 3 From the AM-GM, equality exists at x = y = z = 3 2 . Thus, the minimum value is 3 .