Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).Áp dụng BĐT Cauchy-Schwarz,ta có:
\(=\left(1-\frac{1}{x+1}\right)+\left(1-\frac{1}{y+1}\right)+\left(1-\frac{1}{z+1}\right)\)
\(=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)
Dấu "=" xảy ra khi x = y = z = 1/3
Vậy A min = 3/4 khi x=y=z=1/3
Đầu tiên ta có: 0 < z < 2\(\sqrt{5}\) ⇒ 20−z2 > 0, 3(9−2z) > 0, B−z > 0
\(5x^2+2xyz+4y^2+3z^2=60\)
\(\Leftrightarrow5\left(B-y-z\right)^2+2\left(B-y-z\right)yz+4y^2+3z^2=60\)
\(\Leftrightarrow\left(9-2z\right)y^2-2\left(B-z\right)\left(5-z\right)y+5\left(B-z\right)^2+3\left(z^2-20\right)=0\)
Đế pt theo nghiệm y có nghiệm thì
\(\Delta'=\left(B-z\right)^2\left(5-z\right)^2-\left(9-2z\right)\left(5\left(B-z\right)^2+3\left(z^2-20\right)\right)\ge0\)
\(\Leftrightarrow\left(z^2-20\right)\left(\left(B-z\right)^2-27+6z\right)\ge0\)
\(\Rightarrow\left(B-z\right)^2-27+6z\le0\)
\(\Rightarrow B\le z+\sqrt{27-6z}\le6\)
B đạt Max là 6 khi x = 1; y = 2; z = 3
\(P=\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\ge\dfrac{1}{y}.\dfrac{4}{x+z}=\dfrac{4}{y\left(x+z\right)}\ge\dfrac{4}{\dfrac{\left(y+x+z\right)^2}{4}}=4\)
\(P_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};1;\dfrac{1}{2}\right)\)