\(P=\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\ge\dfrac{1}{y}.\dfrac{4}{x+z}=\dfrac{4}{y\left(x+z\right)}\ge\dfrac{4}{\dfrac{\left(y+x+z\right)^2}{4}}=4\)

\(P_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};1;\dfrac{1}{2}\right)\)

Áp dụng Bất Đẳng Thức Cosi ta có \(\hept{\begin{cases}\frac{x^3}{1+y}+\frac{1+y}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{x^3}{1+y}\cdot\frac{1+y}{4}\cdot\frac{1}{2}}=\frac{3x}{2}\\\frac{y^3}{1+z}+\frac{1+z}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{y^3}{1+z}\cdot\frac{1+z}{4}\cdot\frac{1}{2}}=\frac{3y}{2}\\\frac{z^3}{1+x}+\frac{1+x}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{z^3}{1+x}\cdot\frac{1+x}{4}\cdot\frac{1}{2}}=\frac{3z}{2}\end{cases}}\)

Cộng vế theo vế ta được \(P+\frac{3+x+y+z}{4}+\frac{3}{2}\ge\frac{3}{2}\left(x+y+z\right)\)

\(\Leftrightarrow P\ge\frac{5}{4}\left(x+y+z\right)-\frac{9}{4}\)

Mà ta có \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\ge9\Rightarrow x+y+z\ge3\)

Do đó \(P\ge\frac{5}{4}\cdot3-\frac{9}{4}=\frac{3}{2}\). Dấu "=" xảy ra khi x=y=z=1

Vậy minP=\(\frac{3}{2}\)khi x=y=z=1

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=\(\frac{1}{2}\)

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 \(x=y=z=\frac{1}{2}\)

Không phải ngược đâu nha mọi người,dấu bằng không xảy ra nhé!

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 \(x=y=z=\frac{1}{2}\)

Áp dụng BĐT Cô - si cho 2 số \(\frac{xy}{z};\frac{yz}{x}\)dương ta có: \(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}.\frac{yz}{x}}=2\sqrt{y^2}=2y\)(1)

Tương tự. \(\frac{yz}{x}+\frac{zx}{y}\ge2\sqrt{\frac{yz}{x}.\frac{zx}{y}}=2\sqrt{z^2}=2z\) (2);

\(\frac{xy}{z}+\frac{zx}{y}\ge2\sqrt{\frac{xy}{z}.\frac{zx}{y}}=2\sqrt{x^2}=2x\)(3)

Cộng từng vế của (1)(2)(3) ta được \(2.\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)\ge2\left(x+y+z\right)=2\Rightarrow P\ge1\)

Vậy Min P = 1 tại x= y = z = 1/3

\(xy+yz+zx\le\dfrac{1}{3}\left(x+y+z\right)^2=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)