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8 tháng 3 2017

Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm

\(\Rightarrow\hept{\begin{cases}x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\\\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\ge3\sqrt[3]{\frac{1}{x^3y^3z^3}}=\frac{3}{xyz}\end{cases}}\)

Nhân theo từng vế 

\(\Rightarrow Q\ge3xyz.\frac{3}{xyz}=9\)

Vậy  \(Q_{min}=9\)

Áp dụng bđt AM-GM ta có

\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)

  Ta có   \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng bđt AM-GM ta có

\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)

\(\Rightarrow P\ge6\)

Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)

18 tháng 2 2020

Làm tiếp bài ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ chớ hình như bị ngược dấu ó.Do mình gà nên chỉ biết cô si mù mịt thôi ạ

\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)

\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)

\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)

\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)

Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)

6 tháng 8 2020

Bài này thì AM-GM thôi 

\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)

Sử dụng BĐT AM-GM cho 3 số không âm ta có :

\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)^2}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)

\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)

\(=3\sqrt[3]{\left(\frac{xy}{x}+\frac{1}{x}\right)\left(\frac{yz}{y}+\frac{1}{y}\right)\left(\frac{zx}{z}+\frac{1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Tiếp tục sử dụng AM-GM cho 2 số không âm ta được :

\(3\sqrt[3]{\left(2\sqrt[2]{y\frac{1}{x}}\right)\left(2\sqrt[2]{z\frac{1}{y}}\right)\left(2\sqrt[2]{x\frac{1}{z}}\right)}\ge3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right)\left(2\sqrt{\frac{z}{y}}\right)\left(2\sqrt{\frac{x}{z}}\right)}\)

\(=3\sqrt[3]{8\left(\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}\right)}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)

Vậy \(Min_P=6\)đạt được khi \(x=y=z=\frac{1}{2}\)

24 tháng 5 2020

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18 tháng 8 2020

\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\)

Sử dụng bất đẳng thức AM-GM cho 3 số thực dương ta có : 

\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}.\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}.\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}}\)

\(=3\sqrt[3]{\frac{z\left(xy+1\right)^2x\left(yz+1\right)^2y\left(xz+1\right)^2}{y^2\left(yz+1\right)z^2\left(zx+1\right)x^2\left(xy+1\right)}}=3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)

\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\frac{xy+1}{x}.\frac{yz+1}{y}.\frac{zx+1}{z}}\)

\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Tiếp tục sử dụng BĐT AM-GM cho 2 số thức dương ta có :

\(y+\frac{1}{x}\ge2\sqrt{y\frac{1}{x}}=2\sqrt{\frac{y}{x}}\)

\(z+\frac{1}{y}\ge2\sqrt{z\frac{1}{y}}=2\sqrt{\frac{z}{y}}\)

\(x+\frac{1}{z}\ge2\sqrt{x\frac{1}{z}}=2\sqrt{\frac{x}{z}}\)

Nhân theo vế các bất đẳng thức cùng chiều ta được 

\(\left(y+\frac{1}{x}\right)\left(x+\frac{1}{z}\right)\left(z+\frac{1}{y}\right)\ge8\sqrt{\frac{y}{x}.\frac{x}{z}.\frac{z}{y}}=8\)

Khi đó \(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(x+\frac{1}{z}\right)\left(z+\frac{1}{y}\right)}\ge3\sqrt[3]{8}=3.2=6\)

Dấu = xảy ra khi và chỉ khi \(x=y=z=\frac{1}{3}\)

Vậy MinP=1/3 đạt được khi x=y=z=1/3

21 tháng 1 2017

Áp dụng BĐT Cô - si cho 3 bộ số không âm

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)

\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)

\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)

\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)

\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)

Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)

NV
13 tháng 6 2020

\(\frac{3}{2}\ge x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(P\ge3\sqrt[3]{\frac{x\left(yz+1\right)^2.y\left(zx+1\right)^2.z\left(xy+1\right)^2}{z^2\left(zx+1\right)x^2\left(xy+1\right)y^2\left(yz+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\)

Xét \(Q=\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}=\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{\sqrt{xy}.\sqrt{yz}.\sqrt{zx}}\)

Đặt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c\le\frac{3}{2}\Rightarrow abc\le\frac{1}{8}\)

\(Q=\frac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}=\frac{1+a^2b^2c^2+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2}{abc}\)

\(Q\ge\frac{1+a^2b^2c^2+3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{a^4b^4c^4}}{abc}=\frac{1}{abc}+abc+3\left(\frac{1}{\sqrt[3]{abc}}+\sqrt[3]{abc}\right)\)

\(Q\ge abc+\frac{1}{64abc}+3\left(\sqrt[3]{abc}+\frac{1}{4\sqrt[3]{abc}}\right)+\frac{63}{64abc}+\frac{9}{4\sqrt[3]{abc}}\)

\(Q\ge2\sqrt{\frac{abc}{64abc}}+6\sqrt{\frac{\sqrt[3]{abc}}{4\sqrt[3]{abc}}}+\frac{63}{64.\frac{1}{8}}+\frac{9}{4.\sqrt[3]{\frac{1}{8}}}=\frac{125}{8}\)

\(\Rightarrow P\ge3\sqrt[3]{Q}\ge3\sqrt[3]{\frac{125}{8}}=\frac{15}{2}\)

\(P_{min}=\frac{15}{2}\) khi \(a=b=c=\frac{1}{2}\) hay \(x=y=z=\frac{1}{2}\)

25 tháng 2 2020

\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)\(\Rightarrow\frac{y+z}{x}-1=\frac{z+x}{y}-1=\frac{x+y}{z}-1\)

\(\Rightarrow\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}=\frac{y+z+z+x+x+y}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Do đó:  +) \(\frac{y+z}{x}=2\)\(\Rightarrow y+z=2x\)

+) \(\frac{z+x}{y}=2\)\(\Rightarrow z+x=2y\)

+) \(\frac{x+y}{z}=2\)\(\Rightarrow x+y=2z\)

Ta có: \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{y+x}{y}.\frac{z+y}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=2.2.2=8\)

22 tháng 8 2020

thiếu điều kiện là \(x+y+z\le\frac{3}{2}\)bạn nhớ bổ sung 

Sử dụng bất đẳng thức AM-GM cho 3 số ,ta có :

\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}.\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}.\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}}\)

\(=3\sqrt[3]{\frac{z\left(xy+1\right)^2.x\left(yz+1\right)^2.y\left(xz+1\right)^2}{y^2\left(yz+1\right).z^2\left(zx+1\right).x^2\left(xy+1\right)}}=3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)

\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)

Tiếp tục sử dụng bất đẳng thức AM-GM cho 2 số ,ta được :

\(3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

\(\ge3\sqrt[3]{\left(2\sqrt{y.\frac{1}{x}}\right)\left(2\sqrt{z.\frac{1}{y}}\right)\left(2\sqrt{x.\frac{1}{z}}\right)}=3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right).\left(2\sqrt{\frac{z}{y}}\right).\left(2\sqrt{\frac{x}{z}}\right)}\)

\(=3\sqrt[3]{2.2.2.\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)

Vậy \(P_{min}=6\)đạt được khi \(x=y=z=\frac{1}{2}\)