Maximum & Minimum Values

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willpower
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Maximum & Minimum Values

Unread post by willpower » Wed Nov 23, 2011 11:38 pm

Let x, y and z be real numbers such that x + y + z = 6. Prove that at least one of the numbers (xy + yz), (yz + zx) and (zx + xy) is not greater than 8 and all of them are at most 9.
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

willpower
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Re: Maximum & Minimum Values

Unread post by willpower » Thu Nov 24, 2011 4:58 pm

Got the solution. :D
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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nafistiham
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Re: Maximum & Minimum Values

Unread post by nafistiham » Thu Nov 24, 2011 6:17 pm

why don't you share the solution? :D
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willpower
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Re: Maximum & Minimum Values

Unread post by willpower » Fri Nov 25, 2011 10:38 pm

For proving MINIMUM VALUE
\[x + y + z = 6
\]
\[=> (x + y + z)^2 = 36
\]\[=> x^2 + y^2 + z^2 + 2(xy + xz +yz)= 36\]
\[\frac{1}{2}(x^2 + y^2) + \frac{1}{2}(x^2 + z^2)+ \frac{1}{2}(y^2 + z^2) + 2 (xy + yz + xz) = 36\]

\[x^2 + y^2 \geq 2xy \]
\[=> 3(xy + yz + xz)\leq 36\]
\[=> 2(xy + yz + xz)\leq 24\]
\[=> (xy + yz) + (xy + xz) + (yz + xz)\leq 24\]
If all three numbers were to be equal, each would equal 8. Since their sum cannot be greater than 24, at least one is not greater than 8.
For proving MAXIMUM VALUE:
\[xy + yz = y(x+z)= y (6-y)= 6y - y^2 = - (y - 3)^2 + 9 \]
Hence, the maximum value of the function is 9.
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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