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\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
Theo hệ quả của bất đẳng thức Cauchy - Schwarz
\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)
Mà \(x^2+y^2+z^2\le3\)
\(\Rightarrow xy+yz+xz\le3\)
Ta có \(P=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+xz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow P\ge\frac{\left(1+1+1\right)^2}{xy+1+yz+1+xz+1}=\frac{9}{xy+yz+xz+3}\left(1\right)\)
Ta có : \(xy+yz+xz\le3\)
\(\Rightarrow xy+yz+xz+3\le6\)
\(\Rightarrow\frac{9}{xy+yz+xz+3}\ge\frac{9}{6}=\frac{3}{2}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow P\ge\frac{3}{2}\)
Vậy \(P_{min}=\frac{3}{2}\)
Dấu " = " xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!
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=\(\frac{1}{2}\)
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\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)
\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)
\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)
\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
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 \(x=y=z=\frac{1}{2}\)
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 \(x=y=z=\frac{1}{2}\)
TA CÓ:
\(Q=\frac{x\left(\sqrt{x+zy}-x\right)}{x+yz-x^2}+\frac{y\left(\sqrt{y+zx}-y\right)}{y+zx-y^2}+\frac{z\left(\sqrt{xy+z}-z\right)}{z+xy-z^2}\)
\(=\frac{x\left(\sqrt{x\left(x+y+z\right)+yz}-x\right)}{x\left(x+y+z\right)+yz-x^2}+\frac{y\left(\sqrt{y\left(x+y+z\right)+zx}-y\right)}{y\left(x+y+z\right)-y^2+zx}+\frac{z\left(\sqrt{xy+z\left(x+y+z\right)}-z\right)}{z\left(x+y+z\right)+xy-z^2}\)
\(=\frac{x\left(\sqrt{\left(x+y\right)\left(z+x\right)}-x\right)}{xy+yz+zx}+\frac{y\left(\sqrt{\left(x+y\right)\left(y+z\right)}-y\right)}{xy+yz+zx}+\frac{z\left(\sqrt{\left(y+z\right)\left(z+x\right)}-z\right)}{xy+yz+za}\)
ÁP DỤNG BĐT CÔ-SI TA ĐƯỢC:
\(Q\le\frac{x\left(\frac{x+y+z+x}{2}-x\right)}{xy+zx+yz}+\frac{y\left(\frac{x+y+z+y}{2}-y\right)}{xy+yz+zx}+\frac{z\left(\frac{x+y+z+z}{2}-z\right)}{xy+yz+zx}\)
\(=\frac{xy+zx}{2\left(xy+yz+zx\right)}+\frac{xy+yz}{2\left(xy+yz+zx\right)}+\frac{yz+zx}{2\left(xy+yz+zx\right)}=1\)
DẤU BẰNG XẢY RA \(\Leftrightarrow x=y=z=\frac{1}{3}\)
\(A=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\)
\(\ge\frac{9}{3+x^2+y^2+z^2}\ge\frac{9}{3+3}=\frac{3}{2}\)
đoạn lớn hơn hoặc bằng cụm 9/ (3+xy+yz+zx) ấy, làm sao để có, mình ko hiểu lắm