Cho các số dương x, y thỏa mãn \(x^2+y^2+\frac{1}{xy}=3\). Tìm GTLN của biểu thức: P=\(2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)-\frac{3}{1+2xy}\)
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18 tháng 2 2020
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 \(x=y=z=\frac{1}{2}\)
PN
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 \(x=y=z=\frac{1}{2}\)
PN
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 \(x=y=z=\frac{1}{2}\)
24 tháng 5 2020
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A
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
KK
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
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)