\(B=\dfrac{1}{x^3+y^3}+\dfrac{1}{xy\left(x+y\right)}=\dfrac{1}{x^3+y^3}+\dfrac{3}{3xy\left(x+y\right)}\)

\(B\ge\dfrac{\left(1+\sqrt{3}\right)^2}{x^3+y^3+3xy\left(x+y\right)}=\dfrac{4+2\sqrt{3}}{\left(x+y\right)^3}=4+2\sqrt{3}\)

\(B_{min}=4+2\sqrt{3}\) khi \(\left(x;y\right)=\left(\dfrac{3+\sqrt{3}-\sqrt[4]{12}}{6+2\sqrt{3}};\dfrac{3+\sqrt{3}+\sqrt[4]{12}}{6+2\sqrt{3}}\right)\) và hoán vị

Lời giải:

Áp dụng BĐT Cauchy-Shwarz:

$B=\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{(x+y)^3-3xy(x+y)}+\frac{1}{xy}$

$=\frac{1}{1-3xy}+\frac{1}{xy}=\frac{1}{1-3xy}+\frac{3}{3xy}$

$\geq \frac{(1+\sqrt{3})^2}{1-3xy+3xy}=(1+\sqrt{3})^2$

Vậy $B_{\min}=(1+\sqrt{3})^2$

Dấu "=" xảy ra khi $xy=\frac{1}{2}-\frac{1}{2\sqrt{3}}$

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

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

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

\(6xy=x+y\ge2\sqrt[]{xy}\Rightarrow\sqrt{xy}\ge\dfrac{1}{3}\Rightarrow xy\ge\dfrac{1}{9}\Rightarrow\dfrac{1}{xy}\le9\)

\(M=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{xy+x}{1-xy}+1}{1+\dfrac{xy+x}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{x+1}{1-xy}}{\dfrac{x+1}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{1}{1-xy}+\dfrac{1}{1+xy}}{\dfrac{1}{1-xy}-\dfrac{1}{1+xy}}\)

\(M=\dfrac{1+xy+1-xy}{1+xy-1+xy}=\dfrac{2}{2xy}=\dfrac{1}{xy}\le9\)

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

\(M=3\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)+\dfrac{1}{2xy}\ge\dfrac{12}{2xy+x^2+y^2}+\dfrac{2}{\left(x+y\right)^2}=\dfrac{14}{\left(x+y\right)^2}=14\)

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

Áp dụng bđt đã cho ta có \(M=4\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)-\dfrac{1}{x^2+y^2}\ge\dfrac{16}{2xy+x^2+y^2}-\dfrac{2}{\left(x+y\right)^2}=\dfrac{16}{\left(x+y\right)^2}-\dfrac{2}{\left(x+y\right)^2}=14\).

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