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2)
Theo hệ quả của bất đẳng thức Cauchy ta có
\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
Do \(x^2+y^2+z^2\le3\)
\(\Rightarrow3\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow1\ge xy+yz+xz\)
\(\Rightarrow4\ge xy+yz+xz+3\)
\(\Rightarrow\dfrac{9}{4}\le\dfrac{9}{3+xy+xz+yz}\) ( 1 )
Ta có \(C=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow C=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{9}{3+xy+yz+xz}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow C=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{9}{4}\)
Vậy \(C_{min}=\dfrac{9}{4}\)
Dấu " = " xảy ra khi \(x=y=z=\sqrt{\dfrac{1}{3}}\)
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\(T=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\) ; x + y + z = 1
\(\Rightarrow T=\frac{x+y+z}{16x}+\frac{x+y+z}{4y}+\frac{x+y+z}{z}\)
\(=\frac{1}{16}+\frac{y}{16x}+\frac{z}{16x}+\frac{x}{4y}+\frac{1}{4}+\frac{z}{4y}+\frac{x}{z}+\frac{y}{z}+1\)
\(=\left(\frac{1}{16}+\frac{1}{4}+1\right)+\left(\frac{y}{16x}+\frac{x}{4y}\right)+\left(\frac{z}{16x}+\frac{x}{z}\right)+\left(\frac{z}{4y}+\frac{y}{z}\right)\) (1)
\(x;y;z>0\Rightarrow\frac{y}{16x};\frac{x}{4y};\frac{z}{16x};\frac{x}{z};\frac{z}{4y};\frac{y}{z}>0\)
áp dụng bđt cô si :
\(\frac{y}{16x}+\frac{x}{4y}\ge2\sqrt{\frac{y}{16x}\cdot\frac{x}{4y}}=\frac{1}{4}\) (2)
\(\frac{z}{16x}+\frac{x}{z}\ge2\sqrt{\frac{z}{16x}\cdot\frac{x}{z}}=\frac{1}{2}\) (3)
\(\frac{x}{4y}+\frac{y}{z}\ge2\sqrt{\frac{z}{4y}\cdot\frac{y}{z}}=1\) (4)
(1)(2)(3)(4) \(\Rightarrow T\ge\frac{1}{16}+\frac{1}{4}+1+\frac{1}{4}+\frac{1}{2}+1\)
\(\Rightarrow T\ge\frac{49}{16}\)
dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{y}{16x}=\frac{x}{4y}\\\frac{z}{16x}=\frac{x}{z}\\\frac{z}{4y}=\frac{y}{z}\end{cases}}\Leftrightarrow\hept{\begin{cases}4y^2=16x^2\\z^2=16x^2\\z^2=4y^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=2x\\z=4x\\z=2y\end{cases}}\) có x+y+z = 1
=> x + 2x + 4x = 1
=> x = 1/7
xong tìm ra y = 2/7 và z = 4/7
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2.
Áp dụng bất đẳng thức Cauchy - schwarz ( hay còn gọi là bất đẳng thức Cosi ):
\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}=\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi x = y = z = 1
1:
Áp dụng bất đẳng thức Cô si:
\(x\left(y+\frac{x}{1+y}\right)+y\left(z+\frac{y}{1+z}\right)+z\left(x+\frac{z}{1+x}\right)\)
\(=\left(x+y+z\right)\left[\left(y+\frac{x}{1+y}\right)+\left(z+\frac{y}{1+z}\right)+\left(x+\frac{z}{1+x}\right)\right]\)
\(=1\left[\left(x+y+z\right)+\left(\frac{x}{1+y}+\frac{y}{1+z}+\frac{z}{1+x}\right)\right]\)
\(=1\left[1+\left(\frac{x+y+z}{1+y+1+z+1+x}\right)\right]\)
\(=1\left[1+\left(\frac{1}{3+\left(x+y+z\right)}\right)\right]\)
\(=1\left[1+\frac{1}{4}\right]\)
\(=1+\frac{5}{4}=\frac{9}{4}\)
Dấu "=" xảy ra khi x = y = z = \(\frac{1}{3}\)
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Áp dụng BĐT Cauchy-Schwarz ta có:
`B>=(1+2+3)^2/(x+y+z)=36/6=6`
Dấu "=" xảy ra `<=>(x;y;z)=(3/7;12/7;27/7)`
Vậy `B_(min)=6<=>(x;y;z)=(3/7;12/7;27/7)`
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Lời giải:
Áp dụng BĐT Cô-si:
$\frac{1}{x+1}+\frac{x+1}{4}\geq 1$
$\frac{1}{y+1}+\frac{y+1}{4}\geq 1$
$\frac{1}{1+z}+\frac{1+z}{4}\geq 1$
Cộng theo vế:
$A+\frac{x+y+z+3}{4}\geq 3$
$\Rightarrow A\geq 3-\frac{x+y+z+3}{4}\geq 3-\frac{3+3}{4}=\frac{3}{2}$
Vậy $A_{\min}=\frac{3}{2}$ khi $x=y=z=1$
Dự đoán điểm rơi \(x=y=z=1\)
Khi đó \(\dfrac{1}{1+x}=\dfrac{1}{1+1}=\dfrac{1}{2}\) và \(1+x=1+1=2\)
Ta cần ghép Cô-si \(\dfrac{1}{1+x}\) với \(k\left(1+x\right)\) sao cho đảm bảo đấu "=" xảy ra khi \(x=1\)
Đồng thời khi Cô-si 2 số dương trên thì dấu "=" xảy ra khi \(\dfrac{1}{1+x}=k\left(1+x\right)\Leftrightarrow\dfrac{1}{2}=k.2\Leftrightarrow k=\dfrac{1}{4}\)
Như vậy, áp dụng BĐT Cô-si cho 2 số dương \(\dfrac{1}{1+x}\) và \(\dfrac{1+x}{4}\), ta có \(\dfrac{1}{1+x}+\dfrac{1+x}{4}\ge2\sqrt{\dfrac{1}{1+x}.\dfrac{1+x}{4}}=1\)
Tương tự, ta có \(\dfrac{1}{1+y}+\dfrac{1+y}{4}\ge1\) và \(\dfrac{1}{1+z}+\dfrac{1+z}{4}\ge1\)
Cộng vế theo vế của các BĐT vừa tìm được, ta có \(A+\dfrac{x+y+z+3}{4}\ge3\)\(\Leftrightarrow A\ge3-\dfrac{x+y+z+3}{4}\)
Lại có \(x+y+z\le3\) nên \(A\ge3-\dfrac{x+y+z+3}{4}\Leftrightarrow A\ge3-\dfrac{3+3}{4}=\dfrac{3}{2}\)
Vậy GTNN của A là \(\dfrac{3}{2}\) khi \(x=y=z=1\)
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Sử dụng Cauchy Schwarz và AM - GM ta dễ có:
\(P=x+y+\frac{1}{x}+\frac{1}{y}\ge x+y+\frac{4}{x+y}\)
\(=\left[x+y+\frac{1}{4\left(x+y\right)}\right]+\frac{15}{4\left(x+y\right)}\)
\(\ge2\sqrt{\frac{x+y}{4\left(x+y\right)}}+\frac{15}{4\cdot\frac{1}{2}}=\frac{17}{2}\)
Đẳng thức xảy ra tại x=y=1/4