Maximum & Minimum Values
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
Re: Maximum & Minimum Values
Got the solution.
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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Re: Maximum & Minimum Values
why don't you share the solution?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Maximum & Minimum Values
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.
\[x + y + z = 6
\]
\[=> (x + y + z)^2 = 36
\]\[=> x^2 + y^2 + z^2 + 2(xy + xz +yz)= 36\]
\[\frac{1}{2}(x^2 + y^2) + \frac{1}{2}(x^2 + z^2)+ \frac{1}{2}(y^2 + z^2) + 2 (xy + yz + xz) = 36\]
\[x^2 + y^2 \geq 2xy \]
\[=> 3(xy + yz + xz)\leq 36\]
\[=> 2(xy + yz + xz)\leq 24\]
\[=> (xy + yz) + (xy + xz) + (yz + xz)\leq 24\]
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