Động não tí đi Quỳnh, a thấy bài này cũng không khó.

Bài dễ mừ, có phải Croatia thật ko vậy :))  (viết đề bị nhầm, là x,y,z dương chứ :))

Áp dụng Cauchy-Schwarz dạng cộng mẫu số:

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

\(\frac{\left(x+y+z\right)^2}{\left(x+y\right)\left(x+z\right)+\left(y+z\right)\left(y+x\right)+\left(z+x\right)\left(z+y\right)}=\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

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

Xét \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\Rightarrow\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}\)

\(=\frac{\left(x+y+z\right)^2}{\frac{4}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)

Dấu bằng xảy ra khi và chỉ khi x=y=z,  Xong! :))

Xét \(4\left(x^3+y^3\right)-\left(x+y\right)^3=3\left(x+y\right)\left(x-y\right)^2\ge0\) (Vì x,y > 0)

Suy ra \(z+\sqrt[3]{4\left(x^3+y^3\right)}\ge x+y+z\)

Hay \(\frac{x+y}{z+\sqrt[3]{4\left(x^3+y^3\right)}}\le\frac{x+y}{x+y+z}\)

Tương tự : \(\frac{y+z}{x+\sqrt[3]{4\left(y^3+z^3\right)}}\le\frac{y+z}{x+y+z}\)

\(\frac{z+x}{y+\sqrt[3]{4\left(z^3+x^3\right)}}\le\frac{z+x}{x+y+z}\)

Cộng theo vế được đpcm.

Ta có:

\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{y}{z}+\dfrac{x}{z}+\dfrac{z}{x}\)

Do đó ta chỉ cần chứng minh:

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

Ta có:

\(\dfrac{x}{y}+\dfrac{x}{y}+1\ge3\sqrt[3]{\dfrac{x^2}{y^2}}\) 

Tương tự ...

Cộng lại ta có:

\(2\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\ge3\left(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\right)\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\)

Do đó ta chỉ cần chứng minh:

\(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

\(\Leftrightarrow\left(\sqrt[3]{\dfrac{x}{y}}-\sqrt[3]{\dfrac{x}{z}}\right)^2+\left(\sqrt[3]{\dfrac{y}{x}}-\sqrt[3]{\dfrac{y}{z}}\right)^2+\left(\sqrt[3]{\dfrac{z}{x}}-\sqrt[3]{\dfrac{z}{y}}\right)^2\ge0\) (luôn đúng)

Em thử ạ!Em không chắc đâu.Hơi quá sức em rồi

Ta có: \(VT=\Sigma\frac{x^3}{z+y+yz+1}=\Sigma\frac{x^3}{z+y+\frac{1}{x}+1}\)

\(=\Sigma\frac{x^4}{xz+xy+1+x}=\frac{x^4}{xy+xz+x+1}+\frac{y^4}{yz+xy+y+1}+\frac{z^4}{zx+yz+z+1}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel,suy ra:

\(VT\ge\frac{\left(x^2+y^2+z^2\right)^2}{\left(x+y+z\right)+2\left(xy+yz+zx\right)+3}\)

\(\ge\frac{\left(\frac{1}{3}\left(x+y+z\right)^2\right)^2}{\left(x+y+z\right)+\frac{2}{3}\left(x+y+z\right)^2+3}\) (áp dụng BĐT \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3};ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))

Đặt \(t=x+y+z\ge3\sqrt{xyz}=3\) Dấu "=" xảy ra khi x = y = z

Ta cần chứng minh: \(\frac{\frac{t^4}{9}}{\frac{2}{3}t^2+t+3}\ge\frac{3}{4}\Leftrightarrow\frac{t^4}{9\left(\frac{2}{3}t^2+t+3\right)}=\frac{t^4}{6t^2+9t+27}\ge\frac{3}{4}\)(\(t\ge3\))

Thật vậy,BĐT tương đương với: \(4t^4\ge18t^2+27t+81\)

\(\Leftrightarrow3t^4-18t^2-27t+t^4-81\ge0\)

Ta có: \(VT\ge3t^4-18t^2-27t+3^4-81\)

\(=3t^4-18t^2-27t\).Cần chứng minh\(3t^4-18t^2-27t\ge0\Leftrightarrow3t^4\ge18t^2+27t\)

Thật vậy,chia hai vế cho \(t\ge3\),ta cần chứng minh \(3t^3\ge18t+27\Leftrightarrow3t^3-18t-27\ge0\)

\(\Leftrightarrow3\left(t^3-27\right)-18\left(t-3\right)\ge0\)

\(\Leftrightarrow\left(t-3\right)\left(3t^2+9t+27\right)-18\left(t-3\right)\ge0\)

\(\Leftrightarrow\left(t-3\right)\left(3t^2+9t+9\right)\ge0\)

BĐT hiển nhiên đúng,do \(t\ge3\) và \(3t^2+9t+9=3\left(t+\frac{3}{2}\right)^2+\frac{9}{4}\ge\frac{9}{4}>0\)

Dấu "=" xảy ra khi t = 3 tức là \(\hept{\begin{cases}x=y=z\\xyz=1\end{cases}}\Leftrightarrow x=y=z=1\)

Chứng minh hoàn tất

Em sửa chút cho bài làm ngắn gọn hơn.

Khúc chứng minh: \(4t^4\ge18t^2+27t+81\)

\(\Leftrightarrow4t^4-18t^2-27t-81\ge0\)

\(\Leftrightarrow\left(t-3\right)\left(4t^3+12t^2+18t+27\right)\ge0\)

BĐT hiển nhiên đúng do \(t\ge3\Rightarrow\hept{\begin{cases}t-3\ge0\\4t^3+12t^2+18t+27>0\end{cases}}\)

Còn khúc sau y chang :P Lúc làm rối quá nên không nghĩ ra ạ!

Áp dụng BĐT cosi ta có

\(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+z}{8}+\frac{1+y}{8}\ge\frac{3}{4}x\)

\(\frac{y^3}{\left(1+x\right)\left(1+z\right)}+\frac{1+x}{8}+\frac{1+z}{8}\ge\frac{3}{4}y\)

\(\frac{z^3}{\left(1+y\right)\left(1+x\right)}+\frac{1+y}{8}+\frac{1+x}{8}\ge\frac{3}{4}z\)

Khi đó 

\(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{1}{4}\left(x+y+z\right)-\frac{3}{4}=\frac{1}{2}\left(a+b+c\right)-\frac{3}{4}\)

Mà \(x+y+z\ge3\sqrt[3]{xyz}=3\)

=> \(VT\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)(ĐPCM)

Dấu bằng xảy ra khi x=y=z=1

\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz.\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

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

\(=\frac{1}{2}\left(x+y+z\right)\text{[}\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)\text{]}\)

\(=\frac{1}{2}\left(x+y+z\right)\text{[}\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\text{]}\left(\text{đ}pcm\right)\)

Dùng biến đổi sau: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(VT=z^3+\left(x+y\right)^3-3xy\left(x+y\right)-3xyz\)

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

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]\)

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

\(=VP\)

Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)

Tương tự:

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

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

\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)