tau không biết nhà xin lỗi 

Sử dụng BĐT AM-GM, ta có: 

\(x^3+y^2\ge2yx\sqrt{x}\)

\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2yx\sqrt{x}}=\frac{1}{xy}\)

Tương tự cộng lại suy ra: 

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

Thiếu đk x, y, z nguyên dương phải ko em?

dồn hết về một biến rồi áp dụng cô-si là ra

Theo GT : \(xy+yz+xz=3xyz\Rightarrow\frac{xy+yz+xz}{xyz}=3\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

\(\frac{x^3}{x^2+z}=\frac{x\left(x^2+z\right)}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)

Tương tự , ta có : \(\frac{y^3}{y^2+x}\ge y-\frac{\sqrt{x}}{2}\) ; \(\frac{z^3}{z^2+y}\ge z-\frac{\sqrt{y}}{2}\)

\(\Rightarrow\frac{x^3}{x^2+z}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\)

Vì x ; y ; z dương , áp dụng BĐT Cô - si , ta có :

\(x+1\ge2\sqrt{x};y+1\ge2\sqrt{y};z+1\ge2\sqrt{z}\)

\(\Rightarrow x+y+z+3\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

=> \(\frac{x+y+z+3}{2}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\) => BĐT được c/m

Tiếp tục AD BĐT Cô - si , ta có :

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)

\(\Rightarrow x+y+z\ge\frac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{9}{3}=3\) => BĐT được c/m

Có : \(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{2}\ge\frac{3.3-3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy ...

Lời giải:
Áp dụng BĐT AM-GM ta có:

\(\text{VT}=x-\frac{x}{x^2+z}+y-\frac{y}{y^2+x}+z-\frac{z}{z^2+y}=(x+y+z)-\left(\frac{x}{x^2+z}+\frac{y}{y^2+x}+\frac{z}{z^2+y}\right)\)

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

Từ giả thiết \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

Cauchy-Schwarz:

\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)

\(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\leq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(1+1+1)=9\)

\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)

Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)

Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)

Do đó \(\text{VT}\geq \text{VP}\) (đpcm)

Dấu "=" xảy ra khi $x=y=z=1$

đặt A=\(\frac{1}{x\left(x+1\right)}\) +\(\frac{1}{y\left(y+1\right)}\) +\(\frac{1}{z\left(z+1\right)}\)=\(\frac{1}{x}\)-\(\frac{1}{x+1}\)+\(\frac{1}{y}\)-\(\frac{1}{y+1}\)+\(\frac{1}{z}\)-\(\frac{1}{z+1}\)

Áp dụng BĐT phụ \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\) (bạn tự chứng minh nha,quy đồng ,nhân chéo ,chuyển về )⇒\(\frac{1}{a+b}\) ≤\(\frac{1}{4}\)(\(\frac{1}{a}\)+\(\frac{1}{b}\))

⇒A≥\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)-\(\frac{1}{4}\)(\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)+3)

⇒A≥\(\frac{3}{4}\) (\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\))-\(\frac{3}{4}\)≥\(\frac{3}{4}\) (\(\frac{9}{x+y+z}\))-\(\frac{3}{4}\)

⇒a≥\(\frac{9}{4}\)-\(\frac{3}{4}\)=\(\frac{3}{2}\) dpcm

dấu bằng xảy ra⇔x=y=z=1

BDT CẦN CM TƯƠNG ĐƯƠNG VỚI 

\(x+y+z-\frac{x}{1+y^2}-\frac{y}{1+z^2}-\frac{z}{1+x^2}\le3-\frac{3}{2}=\frac{3}{2}\)

\(\Leftrightarrow\left(x-\frac{x}{1+y^2}\right)+\left(y-\frac{y}{1+z^2}\right)+\left(z-\frac{z}{1+x^2}\right)\le\frac{3}{2}\)

\(\Leftrightarrow\frac{xy^2}{1+y^2}+\frac{yz^2}{1+z^2}+\frac{zx^2}{1+x^2}\le\frac{3}{2}\)(1)

mat khac \(1+y^2\ge2y;1+z^2\ge2z;1+x^2\ge2x\)nen 

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

ma \(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\) (bn tu cm)

\(\Rightarrow vt\le\frac{1}{2}.\frac{\left(x+y+x\right)^2}{3}=\frac{3}{2}\)

dau =xay ra khi va chi khi x=y=z=1

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

Theo AM-GM: \(x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\)

\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\)

Tương tự: \(\frac{2\sqrt{y}}{y^3+z^2}\le\frac{1}{yz}\)

\(\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{zx}\)

Cộng vế với vế => \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

Theo AM-GM; \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}}{2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Dấu " = " xảy ra <=> x=y=z=1

Áp dụng bất đẳng thức Cacuhy - Schwarz 

\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)

Áp dụng bất đẳng thức Cacuchy Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{z^2x^2}}=\frac{2}{xz}\end{cases}}\)

\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

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

Từ (1) và (2)

\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)