Cho \(x,y,z>0\) và \(x\left(x+1\right)+y\left(y+1\right)+z\left(z+1\right)\le18\)
Tìm GTNN của \(P=\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{x+z+1}\)
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Với x,y,z >0 xét gt :
x(x+1) +y(y+1) + z( z+1 ) <=18
<=> ( x^2 + y^2 + z^2 ) + x+ y+z < hoac = 18
áp dụng bdt B.C.S co x^2 + y^2 + z^2 > hoac = ( x+y+z)^2 /3
=> ( x+y+z )^2/3 + (x+y+z) < hoac = 18
dat x+y+z =t ( t > 0)
tu cm dc t nho hon hoac bang 6
áp dụng bdt swarscher vao A => A > hoặc = 9/ ( 2*6 + 1*3 ) = 3/5
Ta có \(x\left(x+1\right)+y\left(y+1\right)+z\left(z+1\right)\le18\)
\(\Leftrightarrow x^2+y^2+z^2+\left(x+y+z\right)\le18\)
\(\Rightarrow54\ge\left(x+y+z\right)^2+3\left(x+y+z\right)\)
\(\Leftrightarrow-9\le x+y+z\le6\)
\(\Leftrightarrow0< x+y+z\le6\)
\(\hept{\begin{cases}\frac{1}{x+y+1}+\frac{x+y+1}{25}\ge\frac{2}{5}\\\frac{1}{y+z+1}+\frac{y+z+1}{25}\ge\frac{2}{5}\\\frac{1}{x+z+1}+\frac{x+z+1}{25}\ge\frac{2}{5}\end{cases}}\)
\(\Rightarrow A+\frac{2\left(x+y+z\right)+3}{25}\ge\frac{6}{5}\Rightarrow A\ge\frac{27}{25}-\frac{2}{25}\left(x+y+z\right)\ge\frac{15}{25}=\frac{3}{5}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=z>0;x+y+z=6\\\left(x+y+1\right)^2=\left(y+z+1\right)^2=\left(z+x+1\right)^2=25\end{cases}\Leftrightarrow x=y=z=2}\)
áp dụng bất đẳng thức Cauchy ta có :
\(\frac{\left(x-1\right)^2}{z}+\frac{z}{4}\ge2\sqrt{\frac{\left(x-1\right)^2}{z}\frac{z}{4}}=|x-1|=1-x.\)
\(\frac{\left(y-1\right)^2}{x}+\frac{x}{4}\ge2\sqrt{\frac{\left(y-1\right)^2}{x}\frac{x}{4}}=|y-1|=1-y.\)
\(\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge2\sqrt{\frac{\left(z-1\right)^2}{y}\frac{y}{4}}=|z-1|=1-z.\)
\(\Rightarrow\frac{\left(x-1\right)^2}{z}+\frac{z}{4}+\frac{\left(y-1\right)^2}{x}+\frac{x}{4}+\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge1-x+1-y+1-z.\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge3-\left(x+y+z\right)-\frac{x+y+z}{4}=3-2-\frac{2}{4}=\frac{1}{2}.\)
Vậy GTNN của \(A=\frac{1}{2}\Leftrightarrow x=y=z=\frac{2}{3}.\)
1. Cho 3 số thực x,y,z thỏa mãn x+y+z=xyz và x,y,z>1
Tìm GTNN của P= x-1/y2 +y-1/x2 + x-1/x2
Giải
Từ gt⇒1xy+1yz+1zx=1⇒1xy+1yz+1zx=1
Theo AM-GM ta có:
P=∑(x−1)+(y−1)y2−∑1y+∑1y2=∑(x−1)(1x2+1y2)−∑1y+∑1y2≥∑(x−1).2xy−∑1y+∑1y2=∑1y+∑1y2−2≥√3∑1xy+∑1xy−2=√3−1P=∑(x−1)+(y−1)y2−∑1y+∑1y2=∑(x−1)(1x2+1y2)−∑1y+∑1y2≥∑(x−1).2xy−∑1y+∑1y2=∑1y+∑1y2−2≥3∑1xy+∑1xy−2=3−1
Dấu = xảy ra⇔x=y=z=1√3
P/S: ĐỀ BÀI TƯƠNG TỰ NÊN BẠN TỰ LÀM NHA !! CHÚC HOK TỐT!
dễ mà bạn :))) gáy tí , sai thì thôi
\(P=\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{y^3}{\left(1+y\right)\left(1+z\right)}+\frac{z^3}{\left(1+z\right)\left(1+x\right)}\)
\(=\frac{x^3\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}+\frac{y^3\left(1+x\right)}{\left(1+y\right)\left(1+x\right)\left(1+z\right)}+\frac{z^3\left(1+y\right)}{\left(1+x\right)\left(1+z\right)\left(1+y\right)}\)
\(=\frac{x^3\left(1+z\right)+y^3\left(1+x\right)+z^3\left(1+y\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\frac{3\sqrt[3]{x^3y^3z^3\left(1+x\right)\left(1+y\right)\left(1+z\right)}}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
đến đây áp dụng BĐT phụ ( 1+a ) ( 1+b ) ( 1+c ) >= 8abc
EZ :)))
a,\(A\ge\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\ge\frac{9}{\sqrt{3\left(x+y+z\right)}}=3\)=3
MInA=3<=>x=y=z=1
b)dùng cô si đi(đề thi chuyên bình phước năm 2016-2017)
\(\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y}{z}\Rightarrow k=2\Rightarrow x=y=z=1\)
A=6
\(\frac{x-y-z}{x}=1-\frac{y+z}{x}\) tương tự con khác
=> x=y=z
=> A=6
ta có
\(0\le\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\left(\forall x,y,z>0\right)\)
\(\Leftrightarrow2xy+2yz+2zx\le2\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\)(1)
dấu = xảy ra khi
\(x=y=z=0\)
theo giả thiết ta có
\(x\left(x+1\right)+y\left(y+1\right)+z\left(z+1\right)\le18\)
\(\Leftrightarrow x^2+y^2+z^2\le18-\left(x+y+z\right)\left(2\right)\)
từ (1) zà (2) suy ra
\(\left(x+y+z\right)^2\le54-3\left(x+y+z\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-54\le0\)
\(\Leftrightarrow\left(x+y+z-6\right)\left(x+y+z+9\right)\le0\)
\(\Leftrightarrow0< x+y+z\le6\left(do\left(x+y+z>0;9>0\right)\right)\)
áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có
\(P=\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\ge\frac{9}{2\left(x+y+z\right)+3}\ge\frac{9}{2.6+3}=\frac{3}{5}\)
Dấu = xảy ra khi zà chỉ khi
\(\hept{\begin{cases}x+y+1=y+z+1=z+x+1\\x+y+z=6\end{cases}=>x=y=z=2}\)
zậy MinP= 3/5 khi x=y=z=2
Ta có : x(x + 1) + y (y+1 ) + z(z + 1) \(\le18\)
<=> x2 + y2 + z2 + ( x + y + z ) \(\le18\)
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
=> 54 \(\ge\)( x + y+z)2 + 3(x + y + z)
<=> -9 \(\le\)x + y + z \(\le\)6
=> 0 \(\le\)x+y+z \(\le\)6
\(\frac{1}{x+y+1}+\frac{x+y+1}{25}\ge\frac{2}{5}\)
\(\frac{1}{y+z+1}+\frac{y+z+1}{25}\ge\frac{2}{5}\)
\(\frac{1}{z+x+1}+\frac{z+x+1}{25}\ge\frac{2}{5}\)
=> \(P+\frac{2\left(x+y+z\right)+3}{25}\ge\frac{6}{5}\)
=> P \(\ge\frac{27}{25}-\frac{2}{25}\left(x+y+z\right)\ge\frac{15}{25}=\frac{3}{5}\)
Dấu " =" xảy ra khi :
\(\hept{\begin{cases}x=y=z>0;x+y+z=6\\\left(x+y+1\right)^2=\left(y+z+1\right)^2=\left(z+x+1\right)^2=25\end{cases}\Leftrightarrow x=y=z=2}\)
Vậy GTNN của P là \(\frac{3}{5}\)khi x = y =z =2