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

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

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 AM-GM:

\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)

\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)

\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

\(minP=3\sqrt{3}\Leftrightarrow x=y=z\)

\(x+\sqrt{3x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{\left(x+y\right)\left(z+x\right)}\ge x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}\)

\(=x+\sqrt{xz}+\sqrt{xy}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+zx}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế với vế ta có đpcm

\(xy+yz+xz=xyz\Rightarrow\)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Áp dụng BĐT Cauchy Schwarz:

\(\dfrac{1}{4x+3y+z}\le\dfrac{1}{64}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

CMTT\(\Rightarrow\) \(M\le\dfrac{1}{64}\left(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\right)=\dfrac{1}{8}\)

Dấu''=" xảy ra\(\Leftrightarrow x=y=z=3\)

Ta chứng minh BĐT sau:

Ta có: \(x\left(3-4x^2\right)=-4x^3+3x-1+1=1-\left(x+1\right)\left(2x-1\right)^2\le1\)

\(\Rightarrow\dfrac{4x^2}{x\left(3-4x^2\right)}\ge\dfrac{4x^2}{1}=4x^2\)

Tương tự và cộng lại:

\(Q\ge4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+zx\right)=3\)

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

Áp dụng bất đẳng thức AM - GM và kết hợp với giả thiết x + y + z = 3 ta có:

\(B=\sqrt{\dfrac{xy}{xy+z\left(x+y+z\right)}}+\sqrt{\dfrac{yz}{yz+x\left(x+y+z\right)}}+\sqrt{\dfrac{zx}{zx+y\left(x+y+z\right)}}\)

\(B=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\dfrac{yz}{\left(y+x\right)\left(z+x\right)}}+\sqrt{\dfrac{zx}{\left(z+y\right)\left(z+x\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}+\dfrac{y}{y+x}+\dfrac{z}{z+x}+\dfrac{z}{z+y}+\dfrac{x}{z+x}\right)\)

\(B\le\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = z = 1.

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