Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}\)
\(\ge\frac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}=\frac{49}{16}\)
Dấu bằng xảy ra khi
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
Ta có: \(A=\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}\)
\(\Rightarrow A+3=\dfrac{x+y}{z}+1+\dfrac{x+z}{y}+1+\dfrac{y+z}{x}+1\)
\(=\dfrac{x+y+z}{z}+\dfrac{x+y+z}{y}+\dfrac{x+y+z}{x}\)
\(=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Mà \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow A+3=0\) \(\Rightarrow A=-3\)
vi a/x + b/y + c/z =0 suy ra ayz/xyz + bxz/xyz + cxy/xyz =0 suy ra ayz+bxz+cxy /xyz =0 suy ra ayz + bxz + cxy =0
vi x/a + y/b =z/c =0 suy ra (x/a + y/b + z/c )^2 =0 suy ra x^2/a^2 +y^2/b^2 + z^2/c^2 + 2(xy/ab + xz/ac + yz/bc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(cxy+ bxz +ayz /abc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 +2011 = 2011
Ta có: \(\dfrac{16}{2x+y+z}\le\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Leftrightarrow\dfrac{1}{2x+y+z}\le\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\left(2\right)\\\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\left(3\right)\end{matrix}\right.\)
Cộng (1), (2), (3) vế theo vế ta được:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{4}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{4.4}{16}=1\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{4}\)
Ta sẽ CM BĐT phụ sau : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Áp dụng BĐT Cauchy dang Engel , ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{3^2}{a+b+c}=\dfrac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Trong đó : \(\left\{{}\begin{matrix}a=x+y\\b=y+z\\c=z+x\end{matrix}\right.\) , ta có :
\(\left(x+y+y+z+x+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)\ge9\)
\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)\ge4,5\)
\(\Leftrightarrow\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}\ge4,5\)
\(\Leftrightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}\ge4,5\)
\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{z+y}\ge1,5\)
\(\Rightarrow P_{Min}=1,5."="\Leftrightarrow x=y=z\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{xy}{z}+\dfrac{yz}{x}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}=2\sqrt{y^2}=2y\left(1\right)\)
\(\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{yz}{x}.\dfrac{xz}{y}}=2\sqrt{z^2}=2z\left(2\right)\)
\(\dfrac{xy}{z}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{xz}{y}}=2\sqrt{x^2}=2x\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3) , ta được :
\(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\) ≥ \(2\left(x+y+z\right)\)
⇔ \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(x+y+z=2019\)
⇒ \(P_{Min}=2019\) ⇔ \(x=y=z=673\)
\(P=\dfrac{\left(x+y\right)\left(y+z\right)}{z+x}+\dfrac{\left(y+z\right)\left(z+x\right)}{x+y}+\dfrac{\left(z+x\right)\left(x+y\right)}{y+z}\)
Áp dụng BĐT Cauchy ta có:
\(\left\{{}\begin{matrix}x+y\ge2\sqrt{xy}\\z+y\ge2\sqrt{yz}\\x+z\ge2\sqrt{xz}\end{matrix}\right.\)
\(\Rightarrow\dfrac{\left(x+y\right)\left(y+z\right)}{z+x}\ge\dfrac{2\sqrt{xy}.2\sqrt{yz}}{2\sqrt{xz}}\)
\(\Leftrightarrow\dfrac{\left(x+y\right)\left(y+z\right)}{z+x}\ge2y\) (1)
Chứng minh tương tự ta có:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\left(y+z\right)\left(z+x\right)}{x+y}\ge2z\left(2\right)\\\dfrac{\left(y+x\right)\left(z+x\right)}{z+y}\ge2x\left(3\right)\end{matrix}\right.\)
Từ (1),(2),(3)
\(\Rightarrow P\ge2x+2y+2z\)
\(\Rightarrow P\ge2.3\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi
\(x=y=z\)
Vậy Min P là 6 khi \(x=y=z\)
Otasaka Yu: Cosi nhưng đừng là ở dưới đó.... (it's same some mô típ i've read and seen Manga and Anime Japan ( ͡° ͜ʖ ͡°))
\(\dfrac{\left(x+y\right)\left(y+z\right)}{x+z}+\dfrac{\left(y+z\right)\left(x+z\right)}{x+y}\ge2\sqrt{\left(y+z\right)^2}=2\left(y+z\right)\)
Tương tự rồi cộng theo vế:
\(2P\ge2\left(x+y+z\right)\Leftrightarrow P\ge x+y+z=3\)
\("=" <=> x=y=z=1\)
It's A jOke. DoN't TriGgeRed my dude !
Đặt a = y + z; b = z+ x; c = x+ y (a;b;c > 0)
=> x+ y + z = (a+b+c)/2
=> x= (a+b+c)/2 - a = (b+c- a)/2
y = (a+b+c)/2 - b = (a+c-b)/2; z = (a+b - c)/ 2
Khi đó \(P=\frac{b+c-a}{2a}+\frac{a+c-b}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}.\left(\frac{b}{a}+\frac{c}{a}-1+\frac{a}{b}+\frac{c}{b}-1+\frac{a}{c}+\frac{b}{c}-1\right)\)
=> \(P=\frac{b+c-a}{2a}+\frac{a+c-b}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}.\left(\left(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)-3\right)\right)\)
AD BĐT Cô - si có: \(\frac{a}{b}+\frac{b}{a}\ge2;\frac{b}{c}+\frac{c}{b}\ge2;\frac{c}{a}+\frac{a}{c}\ge2\)
=> \(P\ge\frac{1}{2}.\left(2+2+2-3\right)=\frac{3}{2}\)=> Min P = 3/2
Dấu "=" khi a = b = c<=> x = y = z
Áp dụng Cauchy - Schwarz và AM-GM :
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{xz+yz}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\)
\(\ge\frac{\left(x+y+z\right)^2}{\frac{2\left(x+y+z\right)^2}{3}}=\frac{3}{2}\)
Đẳng thức xảy ra tại x=y=z
Áp dụng bđt Svác - sơ ta có :
\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{\left(x+y+z\right)^2}{2.\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{3}{2}\) có GTNN là \(\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
@Đinh Đức Hùng Các bất đẳng thức đó các bạn học ở đâu vậy ạ?