làm tương tự bài này nha
x + y + z = 3. Tìm Max P = xy + yz + xz
Ta có: (x - y)² ≥ 0 <=> x² - 2xy + y² ≥ 0 <=> x² + y² ≥ 2xy
hay 2xy ≤ x² + y² , dấu " = " xảy ra <=> x = y
tương tự:
+) 2yz ≤ y² + z² +) 2xz ≤ x² + z²
cộng 3 vế của 3 bđt trên
--> 2xy + 2yz + 2xz ≤ 2(x² + y² + z²)
--> xy + yz + xz ≤ x² + y² + z²
--> xy + yz + xz + 2xy + 2yz + 2xz ≤ x² + y² + z² + 2xy + 2yz + 2xz
--> 3(xy + yz + xz) ≤ (x + y + z)²
--> 3(xy + yz + xz) ≤ 3²
--> xy + yz + xz ≤ 3
Theo đề ta có :
xy + yz + xz = 0
\(\Rightarrow xy=0-yz-xz=-\left(yz+xz\right)\) (1)
\(\Rightarrow yz=0-xz-xy=-\left(xz+xy\right)\)(2)
\(\Rightarrow xz=0-xy-yz=-\left(xy+yz\right)\)(3)
\(M=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
Từ (1) ; (2) và (3) , ta có :
\(M=\frac{-\left(xy+xz\right)}{x^2}+\frac{-\left(xy+yz\right)}{y^2}+\frac{-\left(yz+xz\right)}{z^2}\)
\(M=\frac{-x\left(y+z\right)}{x^2}+\frac{-y\left(x+z\right)}{y^2}+\frac{-z\left(x+y\right)}{z^2}\)
\(M=\frac{-\left(y+z\right)}{x}+\frac{-\left(x+z\right)}{y}+\frac{-\left(x+y\right)}{z}\)
\(M-3=\left(\frac{-\left(y+z\right)}{x}-1\right)+\left(\frac{-\left(x+z\right)}{y}-1\right)+\left(\frac{-\left(x+y\right)}{z}-1\right)\)
\(M-3=\left(\frac{-y-z}{x}-\frac{x}{x}\right)+\left(\frac{-x-z}{y}-\frac{y}{y}\right)+\left(\frac{-x-y}{z}-\frac{z}{z}\right)\)
\(M-3=\left(\frac{-y-z-x}{x}\right)+\left(\frac{-x-z-y}{y}\right)+\left(\frac{-x-y-z}{z}\right)\)
\(M-3=\frac{-\left(y+z+x\right)}{x}+\frac{-\left(x+z+y\right)}{y}+\frac{-\left(x+y+z\right)}{z}\)
..............
\(\frac{xy+xz+yz}{xyz}=0\Rightarrow\frac{1}{z}+\frac{1}{y}+\frac{1}{x}=0\)
voi a+b+c=0 thi \(a^3+b^3+c^3=3abc\)
that vay \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
=\(\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)\)
=0
ap dung ta cung co \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=3\left(\frac{1}{x}.\frac{1}{y}.\frac{1}{z}\right)=\frac{3}{xyz}\)
M=\(\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=0\)
