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

\(=\frac{x^3}{1+z+y+yz}+\frac{y^3}{1+x+z+xz}+\frac{z^3}{1+y+x+xy}\)

\(=\frac{x^3}{1+x+y+2y}\ge\frac{x}{2}\Rightarrow TổngBPT\ge\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\ge\frac{2}{3}\left(đpcm\right)\)

(Không chắc à nha)

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

\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)\)

Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}\)

Từ (1) , (2) và (3) 

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

\(\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}\)

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

Chúc bạn học tốt !!!

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

\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)\)

Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}\)

Từ (1) , (2) , (3)

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

\(\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}\)

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

Chúc bạn học tốt !!!

Áp dụng bđt AM-GM ta có: 

\(\hept{\begin{cases}\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}.\frac{1+y}{8}.\frac{1+z}{8}}=\frac{3x}{4}\left(1\right)\\\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge3\sqrt[3]{\frac{y^3}{\left(1+z\right)\left(1+x\right)}.\frac{1+z}{8}.\frac{1+x}{8}}=\frac{3y}{4}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge3\sqrt[3]{\frac{z^3}{\left(1+x\right)\left(1+y\right)}.\frac{1+x}{8}.\frac{1+y}{8}}=\frac{3z}{4}\left(3\right)\end{cases}}\)

Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\)ta được: 

\(P+\frac{3+x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}\)

\(\Leftrightarrow P\ge\frac{3\left(x+y+z\right)}{4}-\frac{3+x+y+z}{4}\)

\(\Leftrightarrow P\ge\frac{2\left(x+y+z\right)-3}{4}\left(1\right)\)

Áp dụng bdt AM-GM ta có:

\(x+y+z\ge3\sqrt[3]{xyz}=3\)Thay vào (1) ta được:

\(P\ge\frac{2.3-3}{4}\)

\(\Rightarrow P\ge\frac{3}{4}\)Dấu"="xảy ra \(\Leftrightarrow x=y=z\)

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!

a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)

b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)

\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)

Ta có:

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

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

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

Thay vào A được:

\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)

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

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=xy+xz+xy+yz+xz+zy\)

\(=2\left(xy+yz+xz\right)\)

\(=2\)(do xy+yz+xz=1)

=>Đpcm

Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.

dia chi ban vua truy cap khong tim thay

Vì xyz = 1 nên ta có thể đặt \(x=\frac{a^2}{bc};y=\frac{b^2}{ac};z=\frac{c^2}{ab}\left(a,b,c>0,a^2\ne bc,b^2\ne ac,c^2\ne ab\right)\)

Khi đó bất đẳng thức tương đương với

\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ac\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge1\)

Mà ta có

\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ac\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2}\)

Ta cần chứng minh

\(\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2}\ge1\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge0\left(đúng\right)\)

Vậy ta có điều phải chứng minh