Trích:
Nguyên văn bởi man111  Find Max. value of $(a-x)(b-y)(c-z)(ax+by+cz) $. where $a,b,c $ are positive and $(a-x),(b-y) $ and $(c-z) $ are also positive. |
Put $P(x,y,z)=(a-x)(b-y)(c-z)(ax+by+cz) $.
It's obvious that the maximum value of P cannot be satisfying $P_{max} \le 0 $ because P is always greater than 0 where we take any $x,y,z $ such that $0<x<a,0<y<b,0<z<c. $.
By the given condition, $(a-x), (b-y) $ and $(c-z) $ are positive. Then $(ax+by+cz) $ is also positive.
Consequently, just applying AM-GM inequality for 4 positive real numbers system $(a-x, b-y, c-z, ax + by + cz) $ as will be expressing as below, then make both 2 sides to 4th power, we obtain:
$(a - x)(b - y)(c - z)(ax + by + cz) = \frac{1}{{abc}}({a^2} - ax)({b^2} - by)({c^2} - cz)(ax + by + cz) \le \frac{1}{{256abc}}{({a^2} + {b^2} + {c^2})^4} $.
The equality holds if and only if there exists some $(x,y,z) $ satisfies the linear combination: $a - x = b - y = c - z = ax + by + cz $. That system exactly has the only solution because after implying to the linear equations system $\left\{ \begin{array}{l}(a + 1)x + by + cz = a \\x - y = a - b \\ y - z = b - c \\\end{array} \right. $, the determinant $D=a+b+c+1>0 $. Thus, it's not very difficult to compute the appropriate $(x,y,z) $.
Finally, u should check that whether it would be suitable to the given hypothesis or not.
Best regards!
Poincare.
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