Dự đoán dấu "=" khi \(x=y=z=\frac{1}{\sqrt{3}}\Rightarrow S=1\)

Ta chứng minh \(S=1\) là GTNN của \(S\)

Thật vật ta có: \(\frac{1}{4x^2-yz+2}+\frac{1}{4y^2-xz+2}+\frac{1}{4z^2-xy+2}\ge1\)

\(\Leftrightarrow\frac{-4x^2+yz+1}{4x^2-yz+2}+\frac{-4y^2+xz+1}{4y^2-xz+2}+\frac{-4z^2+xy+1}{4z^2-xy+2}\ge0\)

\(\Leftrightarrow\frac{2yz-4x^2+xy+xz}{4x^2-yz+2}+\frac{2xz-4y^2+xy+yz}{4y^2-xz+2}+\frac{2xy-4z^2+xz+yz}{4z^2-xy+2}\ge0\)

\(\LeftrightarrowΣ_{cyc}\frac{-\left(2x+z\right)\left(x-y\right)-\left(2x+y\right)\left(x-z\right)}{4x^2-yz+2}\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(x-y\right)\left(\frac{2y+z}{4y^2-xz+2}-\frac{2x+z}{4x^2-yz+2}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(x-y\right)^2\left(\frac{z^2+6yz+6xz+8xy-4}{\left(4y^2-xz+2\right)\left(4x^2-yz+2\right)}\right)\right)\ge0\) *Đúng*

BĐT cuối đúng hay ta có ĐCPM

bạn có thể trình bày theo bdt cô si hay bunhia  được không

Ta có:

Tương tự ta có: \(\frac{1}{4y^2-zx+2}\ge zx;\frac{1}{4z^2-xy+2}\ge xy\)

Cộng từng vế của 3 bất đẳng thức trên. ta được:

\(\frac{1}{4x^2-yz+2}+\frac{1}{4y^2-zx+2}+\frac{1}{4z^2-xy+2}\ge xy+yz+zx=1\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{\sqrt{3}}{3}\)

Áp dụng BĐT cô si 

\(\frac{xy}{z}+\frac{yz}{x}\ge2y\)

\(\frac{yz}{x}+\frac{xz}{y}\ge2z\)

\(\frac{xz}{y}+\frac{xy}{z}\ge2x\)

Cộng vế với vế của ba BĐT :

=> \(A\ge x+y+z=1\)

Vậy .... 

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{9}{xy+yz+xz}(1)\)

\(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+xz}+\frac{1}{xy+yz+xz}\geq \frac{9}{x^2+y^2+z^2+2(xy+yz+xz)}=\frac{9}{(x+y+z)^2}=9(2)\)

Áp dụng hệ quả quen thuộc của BĐT AM-GM ta có:

\(3(xy+yz+xz)\leq (x+y+z)^2=1\Rightarrow xy+yz+xz\leq \frac{1}{3}\)

\(\Rightarrow \frac{7}{xy+yz+xz}\geq 21(3)\)

Từ \((1);(2);(3)\Rightarrow \frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+xz}\geq 9+21=30\)Vậy $P_{\min}=30$. Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$
​

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{9}{xy+yz+xz}(1)\)

\(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+xz}+\frac{1}{xy+yz+xz}\geq \frac{9}{x^2+y^2+z^2+2(xy+yz+xz)}=\frac{9}{(x+y+z)^2}=9(2)\)

Áp dụng hệ quả quen thuộc của BĐT AM-GM ta có:

\(3(xy+yz+xz)\leq (x+y+z)^2=1\Rightarrow xy+yz+xz\leq \frac{1}{3}\)

\(\Rightarrow \frac{7}{xy+yz+xz}\geq 21(3)\)

Từ \((1);(2);(3)\Rightarrow \frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+xz}\geq 9+21=30\)Vậy $P_{\min}=30$. Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$

Cmr gì bạn 

Ghi đủ đề rùi nhắn tin cho mk biết là đã sửa rùi mk làm cho 

x,y,z>0

bdt trên <=3/2

Do \(0< x;y;z\le1\Rightarrow\left(x-1\right)\left(z-1\right)\ge0\)

\(\Leftrightarrow xz-x-z+1\ge0\)

\(\Leftrightarrow xz+1\ge x+z\Rightarrow1+y+xz\ge x+y+z\)

\(\Rightarrow\frac{x}{1+y+xz}\le\frac{x}{x+y+z}\)

Hoàn toàn tương tự: \(\frac{y}{1+z+xy}\le\frac{y}{x+y+z}\) ; \(\frac{z}{1+x+yz}\le\frac{z}{x+y+z}\)

\(\Rightarrow VT\le\frac{x+y+z}{x+y+z}\le\frac{3}{x+y+z}\) (do \(x;y;z\le1\Rightarrow x+y+z\le3\))

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)

x^2+1>=2x suy ra 1/x^2+1=y<=1/2x+y=1/x+x+y=1/9(9/x+x+y)<=1/x+1/x+1/y.

A(BT)<=1/9(3/x+3/y+3/z)=1/3(1/x+1/y+1/z)

Mà từ x+y+z=xy+yz+zx suy ra x+y+z=xy+yz+zx>=3

dễ dàng cm bằng phương pháp đánh giá suy ra 1/x+1/y+1/z<3

suy ra A<1/3.3=1(đpcm)

\(P=\frac{9}{1-2\left(xy+yz+xz\right)}+\frac{2}{xyz}=\frac{9}{\left(x+y+z\right)^2-2\left(xy+yz+xz\right)}+\frac{2\left(x+y+z\right)}{xyz}\)

\(=\frac{9}{x^2+y^2+z^2}+\frac{6\sqrt[3]{xyz}}{xyz}\ge\frac{9}{x^2+y^2+z^2}+\frac{18}{3\sqrt[3]{x^2y^2z^2}}\)

\(\ge\frac{9}{x^2+y^2+z^2}+\frac{36}{2\left(xy+yx+xz\right)}\ge9\left(\frac{1}{\left(x+y+z\right)^2}+\frac{2^2}{2\left(xy+yz=xz\right)}\right)\)

\(\ge\frac{81}{\left(x+y+z\right)^2=81}\)

Dấu = xảy ra khi x =  y = z = 1/3

Áp dụng BĐT Cô - si cho 3 bộ số không âm

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

Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

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

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

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)

\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)

\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)

Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)

Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)