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\(=\frac{3}{5}\)

1.

\(-1\le a\le2\Rightarrow\hept{\begin{cases}a+1\ge0\\a-2\le0\end{cases}\Rightarrow\left(a+1\right)\left(a-2\right)\le0\Leftrightarrow a^2\le}2+a\)

Tương tự \(b^2\le2+b,c^2\le2+c\Rightarrow a^2+b^2+c^2\le6+a+b+c=6\)

Dấu "=" xảy ra khi a=2,b=c=-1 và các hoán vị của chúng

Xét \(\frac{a^2+1}{a}=a+\frac{1}{a}\)

Dễ thấy dấu "=" xảy ra khi  \(a=\frac{1}{3}\)

khi đó \(a+\frac{1}{a}=a+\frac{1}{9a}+\frac{8}{9a}\ge2\sqrt{\frac{a.1}{9a}}+\frac{8}{\frac{9.1}{3}}=\frac{10}{3}\)

\(\Rightarrow\frac{a}{a^2+1}\le\frac{3}{10}\)

tương tự =>đpcm

A\(\ge3\)

You know

A\(\ge\)9

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

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)