Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)

Theo giả thiết ta có : \(x+yz=yz-z-1=\left(z-1\right)\left(y+1\right)=\left(x+y\right)\left(y+1\right)\)

Tương tự : \(y+zx=\left(x+y\right)\left(x+1\right)\)

Và \(z+xy=\left(x+1\right)\left(y+1\right)\)

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

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

Ta có \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\left(x+1\right)\left(y+1\right)\le\frac{\left(x+y+2\right)^2}{4}\)

nên \(P\ge\frac{2\left(x+y\right)^2+4\left(x+y\right)}{\left(x+y+2\right)^2\left(x+y\right)}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}=\frac{2\left(x+y\right)+4}{\left(x+y+2\right)^2}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}\)

                                                       \(=\frac{2}{z+1}+\frac{4\left(z^2+2\right)}{\left(z+1\right)^2}=f\left(z\right);z>1\)

Lập bảng biến thiên ta được \(f\left(z\right)\ge\frac{13}{4}\) hay min \(P=\frac{13}{4}\) khi \(\begin{cases}z=3\\x=y=1\end{cases}\)

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

\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)

\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)

Do đó:

\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)

Đẳng thức xảy ra khi $x=y=z=1.$

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

Ta xét BĐT phụ: \(1+x^3+y^3\ge xy\left(x+y+z\right)\)

\(x^3+y^3\ge xy\left(x+y\right)+xyz-1\)

\(x^3+y^3-xy\left(x+y\right)\ge0\)

\(\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)

\(\left(x+y\right)\left(x-y\right)^2\ge0\)( Luôn đúng, vậy BĐT phụ đúng)

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

GTNN của P là \(3\sqrt{3}\Leftrightarrow x=y=z=1\)

\(x(3 - xy - xz) + y + 6z \le 5xz(y + z) \quad (1)\)

=>\(3x - x^2 y - x^2 z + y + 6z \le 5xyz + 5xz^2\)

=>\(3x+y+6z\le x^2y+x^2z+5xyz+5xz^2\)

=>\(3x+y+6z\le x^2(y+z)+5xz(y+z)\)

=>\(3x+y+6z\le x(y+z)(x+5z)\quad(2)\)

P=3x+y+6z

=>P<=x(y+z)(x+5z)

y+6z=P-3x

=>y+z=(P-3x)-5z

(2)=>\(P \le x(P - 3x - 5z)(x + 5z)\)

Theo AM-GM, ta có: \((P - 3x - 5z)(x + 5z) \le \left[ \frac{(P - 3x - 5z) + (x + 5z)}{2} \right]^2 = \left( \frac{P - 2x}{2} \right)^2\)

=>\(P \le x \cdot \left( \frac{P - 2x}{2} \right)^2 = \frac{x(P - 2x)^2}{4}\)

=>\(4P\le x(P^2-4Px+4x^2)\)

=>\(4P\le P^2x-4Px^2+4x^3\)

=>\(P^2x-4Px^2-4P+4x^3\ge0\)

=>\(P^2 - 4P \left(x + \frac{1}{x}\right) + 4x^2 \ge 0 \quad (3)\)

Theo Cosi, ta có: \(x + \frac{1}{x} \ge 2\sqrt{x \cdot \frac{1}{x}} = 2\)

=>\(-4P\left(x + \frac{1}{x}\right) \le -4P \cdot 2 = -8P\)

(3)=>\(P^2 - 8P + 4x^2 \ge P^2 - 4P \left(x + \frac{1}{x}\right) + 4x^2 \ge 0\)

Chọn điểm rơi x=1, ta sẽ có phương trình: \(P^2 - 8P + 4 \ge 0\)

=>\(P^2-8P+16\ge12\)

=>\(\left(P-4\right)^2\ge12\)

=>\(P-4\ge2\sqrt3\)

=>\(P\ge4+2\sqrt3\)

Dấu '=' xảy ra khi \(x=\frac{1}{x};P-3x-5z=x+5z\) và y=P-3x-6z

=>x=1; P-4x=10z và y=P-3x-6z

=>x=1; \(10z=(4+2\sqrt{3})-4=2\sqrt{3}\) và y=P-3x-6z

=>x=1; \(z=\frac{\sqrt{3}}{5}\) và y=P-3x-6z

=>\(x=1;y=1+\frac{4\sqrt{3}}{5};z=\frac{\sqrt{3}}{5}\)

Áp dụng bất đẳng thức Cauchy 

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge\frac{9}{xy+yz+zx}\)

\(M\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+xz\right)}+\frac{7}{xy+yz+zx}\)

Áp dụng BĐT Cauchy - Schwarz :

\(\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}\ge\frac{\left(1+2\right)^2}{\left(x+y+z\right)^2}=9\)

và \(\frac{7}{xy+yz+xz}\ge\frac{7}{\frac{1}{3}\left(x+y+z\right)^2}=21\)

\(\Rightarrow M\ge9+21=30\)

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

Áp dụng BĐT Cauchy schwarz ta có:

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

\(\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+zx}\)

\(=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}+\frac{7}{2\left(xy+yz+zx\right)}\)

\(\ge\frac{9}{\left(x+y+z\right)^2}+\frac{7}{\frac{2\left(x+y+z\right)^2}{3}}=30\)

Đẳng thức xảy ra tại x=y=z=¹/₃