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Đặt \(\left\{{}\begin{matrix}x-y=a\\x-z=b\end{matrix}\right.\) \(\Rightarrow ab=1\)
\(S=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a-b\right)^2}=\frac{a^2+b^2}{a^2b^2}+\frac{1}{\left(a-b\right)^2}=a^2+b^2+\frac{1}{\left(a-b\right)^2}\)
\(S=a^2+b^2-2ab+\frac{1}{\left(a-b\right)^2}+2=\left(a-b\right)^2+\frac{1}{\left(a-b\right)^2}+2\)
\(S\ge2\sqrt{\frac{\left(a-b\right)^2}{\left(a-b\right)^2}}+2=4\) (đpcm)
Câu 2:
Từ điều kiện bài này có thể đặt ẩn phụ và AM-GM ra luôn kết quả, nhưng hơi rắc rối khi người ta hỏi từ đâu mà có cách đặt ẩn phụ như vậy, do đó ta giải trâu :D
\(x^2+y^2+z^2+xyz=4\)
\(\Leftrightarrow\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{4}+2\left(\frac{x}{2}.\frac{y}{z}.\frac{z}{2}\right)=1\)
\(\Leftrightarrow\frac{xy}{2z}.\frac{xz}{2y}+\frac{xy}{2z}.\frac{yz}{2x}+\frac{yz}{2x}.\frac{xz}{2y}+2\left(\frac{xy}{2z}.\frac{yz}{2x}.\frac{xy}{2y}\right)=1\)
Đặt \(\left(\frac{xy}{2z};\frac{zx}{2y};\frac{yz}{2x}\right)=\left(m;n;p\right)\Rightarrow mn+np+pn+2mnp=1\)
\(\Leftrightarrow2\left(n+1\right)\left(m+1\right)\left(p+1\right)=\left(n+1\right)\left(m+1\right)+\left(n+1\right)\left(p+1\right)+\left(m+1\right)\left(p+1\right)\)
\(\Leftrightarrow\frac{1}{n+1}+\frac{1}{m+1}+\frac{1}{p+1}=2\)
\(\Leftrightarrow1=\frac{n}{n+1}+\frac{m}{m+1}+\frac{p}{p+1}\ge\frac{\left(\sqrt{n}+\sqrt{m}+\sqrt{p}\right)^2}{m+n+p+3}\)
\(\Leftrightarrow m+m+p+2\left(\sqrt{mn}+\sqrt{np}+\sqrt{mp}\right)\le m+n+p+3\)
\(\Leftrightarrow\sqrt{mn}+\sqrt{np}+\sqrt{mp}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\le\frac{3}{2}\Leftrightarrow x+y+z\le3\)
Câu 1:
\(2xyz=1-\left(x+y+z\right)+xy+yz+zx\)
\(\Rightarrow xy+yz+zx=2xyz+\left(x+y+z\right)-1\)
\(VT=x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\)
\(=\left(x+y+z\right)^2-2\left(x+y+z\right)-4xyz+2\)
\(VT\ge\left(x+y+z\right)^2-2\left(x+y+z\right)-\frac{4}{27}\left(x+y+z\right)^3+2\)
\(VT\ge\frac{4}{27}\left[\frac{15}{4}-\left(x+y+z\right)\right]\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}\)
(Do \(0< x;y;z< 1\Rightarrow x+y+z< 3< \frac{15}{4}\))
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
Ta có:
\(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)
\(\Leftrightarrow\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}=-2\)
\(\Leftrightarrow x^2z+x^2y+y^2x+y^2z+z^2x+z^2y+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=-z\\z=-x\end{cases}}\)
Với \(x=-y\)
\(\Rightarrow x^3+y^3+z^3=1\)
\(\Rightarrow z=1\)
\(\Rightarrow P=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x}+\frac{1}{-x}+\frac{1}{1}=1\)
Tương tự cho các trường hợp còn lại.
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{2x+y}{8}+\frac{y+z}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\\\frac{y^3}{\left(2y+z\right)\left(z+x\right)}+\frac{2y+z}{8}+\frac{x+z}{8}\ge3\sqrt[3]{\frac{y^3}{64}}=\frac{3y}{4}\\\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{2z+x}{8}+\frac{x+y}{8}\ge3\sqrt[3]{\frac{z^3}{64}}=\frac{3z}{4}\end{cases}}\)
\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{5\left(x+y+z\right)}{8}\ge\frac{3\left(x+y+z\right)}{4}\)
\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{5}{8}\ge\frac{3}{4}\)
\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}\ge\frac{1}{8}\)
\(\Leftrightarrow P_{min}=\frac{1}{8}\)
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
Sao không nói x , y , z thuộc N cho nhanh bạn