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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\(=\frac{3}{5}\)
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
\(\frac{a}{1+2b^3}=\frac{a\left(1+2b^3\right)-2ab^3}{1+2b^3}=a-\frac{2ab^3}{1+2b^3}\)
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
\(a+bc=a\left(a+b+c\right)+bc=a^2+ab+ac+bc=\left(a+b\right)\left(a+c\right)\)
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
\(P\le1\)
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)