Đặt \(a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\),xyz=1  

Cần CM: \(1+\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\ge\frac{6}{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}}\) 

\(\Leftrightarrow1+\frac{3}{xy+yz+zx}\ge\frac{6}{x+y+z}\) 

Thật vậy \(1+\frac{3}{xy+yz+zx}\ge1+\frac{9}{\left(x+y+z\right)^2}\ge2\sqrt{\frac{9}{x+y+z}}=\frac{6}{x+y+z}\)(đpcm) 

Dấu "=" xảy ra khi a=b=c=1

Bài 2:

\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)

\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)

\(\Rightarrow P\ge\sqrt[3]{3}\)

Dấu bằng xẩy ra khi a=b=c=3

Bài 1: 

 \(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)

Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)

\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)

Áp dụng bđt AM-GM ta có:

 \(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)

\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)

\(\Rightarrow\)(*) luôn đúng

Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)

Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)

Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)

\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)

https://olm.vn/thanhvien/ankhunge

Làm sai rồi ạ

khó quá nha bn

mk mới chỉ hok lớp 7 thôi

xin lỡi nha

mk tin sẽ có nguoi tra lới cau hoi của bn

hok tot >_<

MIN=1=>a=b=c=1

ta có 

\(\frac{a}{1+2b^3}=\frac{a\left(1+2b^3\right)-2ab^3}{1+2b^3}=a-\frac{2ab^3}{1+2b^3}\)

Vì \(1+2b^3\ge3b^2\left(cosi\right)\)

\(\Rightarrow a-\frac{2ab^3}{a+2b^3}\ge a-\frac{2}{3}ab\)

cmtt ta đc 

P\(\ge a+b+c-\frac{2}{3}\left(ab+bc+ca\right)\)

\(P\ge a+b+c-2\)

mặt khác \(\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ca\)

\(\Rightarrow a+b+c\ge3\)

\(\Rightarrow P\ge3-2=1\)

Dấu = xảy ra a=b=c=1

Ta có: \(a^2+b^2\ge2ab\forall a,b\Rightarrow\frac{1}{4-ab}\le\frac{2}{8-a^2-b^2}\)

Theo BĐT C-S: \(\frac{2}{8-a^2-b^2}\le\frac{1}{2}\left(\frac{1}{4-a^2}+\frac{1}{4-b^2}\right)\)

Do đó: \(\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca}\le\frac{1}{4-a^2}+\frac{1}{4-b^2}+\frac{1}{4-c^2}\)

Ta có đánh giá sau: \(\frac{1}{4-a^2}\le\frac{a^4+5}{18}\Leftrightarrow\left(a^2-1\right)^2\left(a^2-2\right)\le0\) (Đúng)

Thiết lập các BĐT tương tự rồi cộng theo vế ta có: 

\(\frac{1}{4-a^2}+\frac{1}{4-b^2}+\frac{1}{4-c^2}\le\frac{a^4+5}{18}+\frac{b^4+5}{18}+\frac{c^4+5}{18}=1\)(ĐPCM)

Đẳng thức xảy ra khi \(a=b=c=1\)

Cách khác dùng Schur như sau :)

BĐT cần chứng minh tương đương với:

\(16+3abc\left(a+b+c\right)\ge a^2b^2c^2+8\left(ab+bc+ca\right)\)

Mà \(1\ge a^2b^2c^2\). Mặt khác theo BĐT Schur ta có: 

\(\left(a^3+b^3+c^3+3abc\right)\left(a+b+c\right)\ge\)

\(\ge\left[ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\right]\left(a+b+c\right)\)

\(\Leftrightarrow3+3abc\left(a+b+c\right)\ge2\left(a^2b^2+b^2c^2+c^2a^2\right)+2abc\left(a+b+c\right)\)

\(=\left(ad+bc\right)^2+\left(bc+ca\right)^2+\left(ca+ab\right)^2\)

BĐT sẽ được c/m xong nếu ta chỉ ra: 

\(\left(ab+bc\right)^2+\left(bc+ca\right)^2+\left(ca+ab\right)^2+12\ge8\left(ab+bc+ac\right)\) 

Đúng theo BĐT Cô-si

Dấu đẳng thức xảy ra khi \(a=b=c=1\)

\(a+bc=a\left(a+b+c\right)+bc=a^2+ab+ac+bc=\left(a+b\right)\left(a+c\right)\)

tương tự :

\(b+ac=\left(b+a\right)\left(b+c\right);c+ba=\left(b+c\right)\left(c+a\right)\)

\(P=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)

áp dụng bất đẳng thức cauchy cho hai số dương 

\(\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)

\(\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{b}{b+c}+\frac{b}{b+a}\right)\)

\(\frac{c}{\sqrt{\left(c+b\right)\left(c+a\right)}}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{c}{c+a}\right)\)

cộng vế theo vế

\(P\le1\)

\(P\le\frac{3}{2}\)

Ta có :\(\frac{1}{a^2+b^2+c^2}+\frac{2009}{ab+bc+ca}\)

\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{2007}{ab+bc+ca}\)

Áp dụng bđt Cauchy - Schwarz dạng Engel ta có : 

\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)

\(=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{3^2}=1\)( do \(a+b+c\le3\)) (1)

Lại có : \(a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)

nên \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)\ge3\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\Leftrightarrow9\ge3\left(ab+bc+ac\right)\Rightarrow ab+bc+ac\le3\)

\(\Rightarrow\frac{2007}{ab+bc+ac}\ge\frac{2007}{3}=669\)(2)

Từ (1) ; (2) \(\Rightarrow\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{2007}{ab+bc+ca}\ge670\)

Hay \(\frac{1}{a^2+b^2+c^2}+\frac{2009}{ab+bc+ca}\ge670\)(đpcm)