Áp dụng bất đẳng thức Côsi cho các số dương $x, y, z$, ta được:$x^{3}+y^{2} \geqslant 2 \sqrt{x^{3} \cdot y^{2}}=2 x y \cdot \sqrt{x}$$y^{3}+z^{2} \geqslant 2 \sqrt{y^{3} \cdot z^{2}}=2 y z \cdot \sqrt{y}$$z^{3}+x^{2} \geqslant 2 \sqrt{z^{3} \cdot x^{2}}=2 z x \cdot \sqrt{z}$Khi đó BĐT đã cho trở thành:$\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{2 \sqrt{x}}{2 x y \sqrt{x}}+\dfrac{2 \sqrt{y}}{2 y z \sqrt{y}}+\dfrac{2 \sqrt{z}}{2 z x \sqrt{z}}=\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} (1)$Mặt khác ta có:$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}} \geqslant \dfrac{2}{x y} \Rightarrow \dfrac{1}{x y} \leqslant \dfrac{1}{2}\left(\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}\right)$

CMTT: $\dfrac{1}{y z} \leq \dfrac{1}{2}\left(\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}\right) ; \dfrac{1}{z x} \leqslant \dfrac{1}{2}\left(\dfrac{1}{z^{2}}+\dfrac{1}{x^{2}}\right)$Suy ra: $\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}(2)$Từ  $(1)$ và $(2)$ ta được: $\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}$Dấu " $="$ xảy ra $\Leftrightarrow x=y=z=1$

Áp dụng bất đẳng thức Côsi cho các số dương $x, y, z$, ta được:

$x^{3}+y^{2} \geqslant 2 \sqrt{x^{3} \cdot y^{2}}=2 x y \cdot \sqrt{x}$

$y^{3}+z^{2} \geqslant 2 \sqrt{y^{3} \cdot z^{2}}=2 y z \cdot \sqrt{y}$

$z^{3}+x^{2} \geqslant 2 \sqrt{z^{3} \cdot x^{2}}=2 z x \cdot \sqrt{z}$

Khi đó BĐT đã cho trở thành:

$\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{2 \sqrt{x}}{2 x y \sqrt{x}}+\dfrac{2 \sqrt{y}}{2 y z \sqrt{y}}+\dfrac{2 \sqrt{z}}{2 z x \sqrt{z}}=\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} (1)$

Mặt khác ta có:

$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}} \geqslant \dfrac{2}{x y} \Rightarrow \dfrac{1}{x y} \leqslant \dfrac{1}{2}\left(\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}\right)$

CMTT: $\dfrac{1}{y z} \leq \dfrac{1}{2}\left(\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}\right) ; \dfrac{1}{z x} \leqslant \dfrac{1}{2}\left(\dfrac{1}{z^{2}}+\dfrac{1}{x^{2}}\right)$

Suy ra: $\dfrac{1}{x y}+\dfrac{1}{y z}+\dfrac{1}{z x} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}(2)$

Từ  $(1)$ và $(2)$ ta được: $\dfrac{2 \sqrt{x}}{x^{3}+y^{2}}+\dfrac{2 \sqrt{y}}{y^{3}+z^{2}}+\dfrac{2 \sqrt{z}}{z^{3}+x^{2}} \leqslant \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}+\dfrac{1}{z^{2}}$

Dấu " $="$ xảy ra $\Leftrightarrow x=y=z=1$

theo bđt cauchy schwarz ta có

\(\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2\sqrt{y^3z^2}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2\sqrt{z^3y^2}}=\dfrac{1}{zy}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}+\dfrac{\dfrac{1}{y^2}+\dfrac{1}{z^2}}{2}+\dfrac{\dfrac{1}{z^2}+\dfrac{1}{x^2}}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)\(\Rightarrow dpcm\)

Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa

Đặt \(\left(\dfrac{1}{\sqrt{x}};\dfrac{1}{\sqrt{y}};\dfrac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}=1\)

Ta cần chứng minh: \(ab+bc+ca\le\dfrac{3}{2}\)

Thật vậy, ta có:

\(1=\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}\)

\(\Rightarrow a^2+b^2+c^2+3\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca\le\dfrac{3}{2}\) (đpcm)

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

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow ab+bc+ca=1\)

Đặt vế trái là P, ta có:

\(P=\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\)

\(P=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)

\(P=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)+\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)

\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)

=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)

=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)

Dấu = xảy ra khi x=y=z=6căn 2