v~~ ko thằng admin :(( t làm cái bài này mất gần 30 phút mà bây giờ nó éo hiện câu trả lời của tao ???? hận quá đi 

bài này easy lắm bạn ơi :(( 

áp dụng BDT (Am-ag) mẫu ta có

\(\left(x^2+y^2\right)\ge2\sqrt{x^2y^2}=2xy\) rồi thay vào

suy ra   \(\frac{1}{x^2+y^2+2}\le\frac{1}{2xy+2}\)

\(\left(y^2+z^2\right)\ge2yz\)

suy ra \(\frac{1}{y^2+z^2+2}\le\frac{1}{2yz+2}\)

tượng tự vs  BDT con lại rồi + vế vs vế ta được

\(VT\le\frac{1}{2xy+2}+\frac{1}{2yz+2}+\frac{1}{2xz+2}=\frac{1}{xy+xy+1+1}+\frac{1}{yz+yz+1+1}+\frac{1}{xz+xz+1+1}\)

gọi cái  \(\frac{1}{yz+yz+1+1}+.........=Pain\)

áp dụng cosi sáp cho 4 số ta được

\(\frac{1}{xy+xy+1+1}\le\frac{1}{16}\left(\frac{1}{xy}+\frac{1}{xy}+\frac{1}{1}+\frac{1}{1}\right)\)

\(\frac{1}{yz+yz+1+1}\le\frac{1}{16}\left(\frac{1}{yz}+\frac{1}{yz}+\frac{1}{1}+\frac{1}{1}\right)\)

\(\frac{1}{xz+xz+1+1}\le\frac{1}{16}\left(\frac{1}{xz}+\frac{1}{xz}+\frac{1}{1}+\frac{1}{1}\right)\)

+ vế với vế ta được

\(VT\le Pain\le\frac{1}{16}\left(\frac{2}{xz}+\frac{2}{yz}+\frac{2}{xy}+\frac{2}{2}+\frac{2}{2}+\frac{2}{2}\right)\)

\(VT\le PAIN\le\frac{1}{8}\left(\frac{1}{xz}+\frac{1}{yz}+\frac{1}{xy}+1+1+1\right)\)

bây giờ m đi chứng minh cái \(\frac{1}{zy}+\frac{1}{yz}+\frac{1}{xy}\ge3\) chắc là m làm được

áp dụng BDT cô si ta có

\(\frac{1}{xz}+xz\ge2\)

\(\frac{1}{yz}+yz\ge2\)

\(\frac{1}{xz}+zx\ge2\)

+ vế với vế ta được

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+xy+yz+zx\ge6\)

mà đề bài cho xy+yz+xz=3 suy ra

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge3\)

nhưng mà nó trái dấu oy :(( kệ nhé cứ thay vào nhé không sao hết bạn oy :)

thay vào ta được

\(VT\le PAIN\le\frac{1}{8}\left(3+3\right)=\frac{3}{4}\)

ĐIỀU CẦN PHẢI CHỨNG MINH :(( 

dù sai dấu nhưng mình vẫn k :)

Lời giải:

Ta có:

\(\text{VT}=\frac{1}{x^2+y^2+2}+\frac{1}{y^2+z^2+2}+\frac{1}{z^2+x^2+2}\)

\(\Rightarrow 2\text{VT}=\frac{2}{x^2+y^2+2}+\frac{2}{y^2+z^2+2}+\frac{2}{z^2+x^2+2}\)

\(2\text{VT}=1-\frac{x^2+y^2}{x^2+y^2+2}+1-\frac{y^2+z^2}{y^2+z^2+2}+1-\frac{z^2+x^2}{z^2+x^2+2}\)

\(2\text{VT}=3-\left(\frac{x^2+y^2}{x^2+y^2+2}+\frac{y^2+z^2}{y^2+z^2+2}+\frac{z^2+x^2}{z^2+x^2+2}\right)=3-A\)

Áp dụng BĐT Cauchy-Schwarz:

\(A\geq \frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2)+6}=\frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2+xy+yz+xz)}(*)\)

Xét tử số:

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

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

Áp dụng BĐT Bunhiacopxky:

\(\sqrt{(x^2+y^2)(x^2+z^2)}\geq \sqrt{(x^2+yz)^2}=x^2+yz\)

\(\sqrt{(x^2+y^2)(y^2+z^2)}\geq \sqrt{(xz+y^2)^2}=xz+y^2\)

\(\sqrt{(y^2+z^2)(z^2+x^2)}\geq \sqrt{(z^2+xy)^2}=z^2+xy\)

\(\Rightarrow \sum \sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+y^2+z^2+xy+yz+xz\)

\(\Rightarrow (\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\geq 4(x^2+y^2+z^2)+2(xy+yz+xz)\)

\(\geq 3(x^2+y^2+z^2)+3(xy+yz+xz)=3(x^2+y^2+z^2+xy+yz+xz)\)

(theo BĐT AM-GM)

Do đó: Từ \((*)\Rightarrow A\geq \frac{3(x^2+y^2+z^2+xy+yz+xz)}{2(x^2+y^2+z^2+xy+yz+xz)}=\frac{3}{2}\)

\(\Rightarrow 2\text{VT}\leq 3-\frac{3}{2}=\frac{3}{2}\)

\(\Rightarrow \text{VT}\leq \frac{3}{4}\) (đpcm)

Dấu bằng xảy ra khi \(x=y=z=1\)

We have: \(\dfrac{1}{x^2+y^2+2}=\dfrac{1}{x^2+y^2+z^2+2-z^2}\le\dfrac{1}{5-z^2}\)

Similarly and by adding them:

\(\dfrac{1}{5-x^2}+\dfrac{1}{5-y^2}+\dfrac{1}{5-z^2}\le\dfrac{3}{4}\left(\circledast\right)\)

We know that \(\dfrac{1}{5-x^2}\le\dfrac{3\left(x^2+x\right)}{8\left(x^2+x+1\right)}\)

\(\Leftrightarrow-\dfrac{\left(x-1\right)^2\left(3x^2+9x+8\right)}{8\left(x^2-5\right)\left(x^2+x+1\right)}\le0\) It's obviously

\(\Rightarrow L.H.S_{\left(\circledast\right)}\le\dfrac{3}{8}\left(\dfrac{x^2+x}{x^2+x+1}+\dfrac{y^2+y}{y^2+y+1}+\dfrac{z^2+z}{z^2+z+1}\right)\le\dfrac{3}{4}\)

The equality occur when \(x=y=z=1\)

Done!

Áp dụng BĐT AM-GM cho 3 số không âm, ta có: \(0< \sqrt[3]{yz.1}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{3x}{y+z+1}\)

Làm tương tự với 2 hạng tử còn lại rồi cộng theo vế thì có:

\(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{z+x+1}+\frac{z}{x+y+1}\right)\)

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

\(=3.\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x+y+z+2\left(xy+yz+zx\right)}\ge9.\frac{xy+yz+zx}{\sqrt{3\left(x^2+y^2+z^2\right)}+2\left(x^2+y^2+z^2\right)}\)

\(=9.\frac{xy+yz+zx}{3+2.3}=xy+yz+zx\) => ĐPCM.

Dấu "=" xảy ra khi x=y=z=1.

\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)

\(\Rightarrow xyz\le1\)

\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)

Ta co:

\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)

\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)

\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)

\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow A\ge xy+yz+zx\)

Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))

Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)

\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)

\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)

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

\(\ge xy+yz+zx\)

Đẳng thức xảy ra khi x = y = z = 1

\(\sqrt[3]{yz\cdot1}\le\frac{y+z+1}{3};\sqrt[3]{xz\cdot1}\le\frac{x+z+1}{3};\sqrt[3]{yx\cdot1}\le\frac{y+x+1}{3}\)

Nên \(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{y+x+1}\right)\)\(=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)=B\)

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

do \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3=x^2+y^2+z^2;xy+yz+zx\le x^2+y^2+z^2=3\)

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

\(=\frac{2x}{3+x^2}+\frac{2y}{3+y^2}+\frac{2z}{3+z^2}\le\frac{2x}{4\sqrt[4]{x^2}}+\frac{2y}{4\sqrt[4]{y^2}}+\frac{2z}{4\sqrt[4]{z^2}}\)

\(=\frac{\sqrt{x}}{2}+\frac{\sqrt{y}}{2}+\frac{\sqrt{z}}{2}\le\frac{x^2+3}{8}+\frac{y^2+3}{8}+\frac{z^2+3}{8}\)

\(=\frac{3}{8}+\frac{9}{8}=\frac{3}{2}\)

cách khác: cũng đến chỗ <= sigma 2x/3+x^2 

<= 2x/2(x+1) (do x^2+3=x^2+1+2>=2x+2) <= sigma x/x+1 = 3- sigma (1/x+1) 

sigma 1/x+1 >= 9/x+y+z+3 dễ rồi

ta caàn chứng minh bđt 

\(\frac{x}{x+yz}+\frac{y}{y+zx}\ge\frac{x}{x+xz}+\frac{y}{y+yz}=\frac{1}{1+z}+\frac{1}{1+z}=\frac{2}{1+z}\)

tương tự + vào, dùng svác sơ

Áp dụng BĐT Cauchy-Schwarz ta có: 

\(VT=\frac{2x^2+y^2+z^2}{4-yz}+\frac{2y^2+z^2+x^2}{4-xz}+\frac{2z^2+x^2+y^2}{4-xy}\)

\(\ge\frac{4x\sqrt{yz}}{4-yz}+\frac{4y\sqrt{xz}}{4-xz}+\frac{4z\sqrt{xy}}{4-xy}\)

Cần chứng minh \(\frac{4x\sqrt{yz}}{4-yz}+\frac{4y\sqrt{xz}}{4-xz}+\frac{4z\sqrt{xy}}{4-xy}\ge4xyz\)

\(\Leftrightarrow\frac{\sqrt{yz}}{yz\left(4-yz\right)}+\frac{\sqrt{xz}}{xz\left(4-xz\right)}+\frac{\sqrt{xy}}{xy\left(4-xy\right)}\ge1\)

Cauchy-Schwarz: \(\left(x+y+z\right)^2\ge\left(1+1+1\right)\left(xy+yz+xz\right)\ge\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)^2\)

\(\Leftrightarrow3\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)

Đặt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{xz}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\hept{\begin{cases}a,b,c>0\\a+b+c\le3\end{cases}}\)

\(\Leftrightarrow\frac{a}{a^2\left(4-a^2\right)}+\frac{b}{b^2\left(4-b^2\right)}+\frac{c}{c\left(4-c^2\right)}\ge1\left(\odot\right)\)

Ta có BĐT phụ: \(\dfrac{a}{a^2\left(4-a^2\right)}\le-\dfrac{1}{9}a+\dfrac{4}{9}\)

\(\Leftrightarrow\dfrac{\left(a-1\right)^2\left(a^2-2a-9\right)}{9a\left(a-2\right)\left(a+2\right)}\le0\forall0< a\le1\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế

\(VT_{\left(\odot\right)}\ge\dfrac{-\left(a+b+c\right)}{9}+\dfrac{4}{9}\cdot3\ge\dfrac{-3}{9}+\dfrac{12}{9}=1=VP_{\left(\odot\right)}\)

Dấu "=" <=> x=y=z=1

em là pô pô nê người con của Thái Nguyên

Bài này có nhiều cách làm. Cách khác:

Gọi vế trái của BĐT là P. Khi đó biến đổi P như sau:

\(P=\left(\frac{x^2}{4-yz}+\frac{y^2}{4-xz}+\frac{z^2}{4-yx}\right)+\left(x^2+y^2+z^2\right)\left(\frac{1}{4-yz}+\frac{1}{4-xz}+\frac{1}{4-yx}\right)\)

Theo BĐT Bunhiacopsky dạng phân thức ta có:

\(\frac{x^2}{4-yz}+\frac{y^2}{4-xz}+\frac{z^2}{4-yx}\ge\frac{\left(x+y+z\right)^2}{12-\left(xy+yz+zx\right)}\)

\(\frac{1}{4-yz}+\frac{1}{4-xz}+\frac{1}{4-yx}\ge\frac{9}{12-\left(xy+yz+zx\right)}\)

Do đó ta được:

\(P\ge\frac{\left(x+y+z\right)^2}{12-\left(xy+yz+xz\right)}+\frac{9\left(x^2+y^2+z^2\right)}{12-\left(xy+yz+xz\right)}\)

\(\ge\frac{3\left(xy+yz+xz\right)}{12-\left(xy+yz+xz\right)}+\frac{9\left(xy+yz+xz\right)}{12-\left(xy+yz+xz\right)}\)

\(\ge\frac{12\left(xy+yz+xz\right)}{12-\left(xy+yz+zx\right)}\ge\frac{36\sqrt[3]{x^2y^2z^2}}{12-3\sqrt[3]{x^2y^2z^2}}\)

đặt \(\sqrt[3]{xyz}=t\le\frac{x+y+z}{3}=1\). Khi đó ta có:

\(\frac{36t^2}{12-3t^2}-4t^3\Leftrightarrow12t^2\left(t-1\right)\left(t^2+t-3\right)\ge0\)

Đánh giá BĐT cuối cùng luôn đúng. BĐT được chứng minh xong