chia thành 3 TH:
a=b
p | a 
p không chia hết a.
GL

Đặt \(ab=x\); \(bc=y\);\(ac=z\)

\(BPT< =>\left(x+y+z\right)^2\ge3\left(xz+xy+yz\right)\)

\(< =>x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\)

\(< =>x^2+y^2+z^2-xy-xz-yz\ge0\)

\(< =>2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)

\(< =>\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\left(LĐ\right)\)

\(a^2+b^2+c^2\ge2\left(ab+bc+ac\right)=2\times9=18\)
 

Nguyễn Quỳnh Anh Sai rồi nhé!

BĐT Cô-si đê ông

\(2.\sqrt{a}+3.\sqrt[3]{b}+4.\sqrt[4]{c}\)

\(=\sqrt{a}+\sqrt{a}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[4]{c}+\sqrt[4]{c}+\sqrt[4]{c}+\sqrt[4]{c}\)

Áp dụng BĐT AM-GM ta có:

\(2.\sqrt{a}+3.\sqrt[3]{b}+4.\sqrt[4]{c}\ge9\sqrt[9]{\sqrt{a}.\sqrt{a}.\sqrt[3]{b}.\sqrt[3]{b}.\sqrt[3]{b}.\sqrt[4]{c}.\sqrt[4]{c}.\sqrt[4]{c}.\sqrt[4]{c}}=9.\sqrt[9]{abc}\)

                                                                                                                                           đpcm    

Áp dụng BĐT AM-GM 

Ta có: \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{3}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

         \(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\ge\frac{3\sqrt[3]{abc}}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

Cộng vế theo vế ta được:

\(\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}+\frac{1}{c+1}+\frac{c}{c+1}\ge\frac{3+3\sqrt[3]{abc}}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

\(\Leftrightarrow1+1+1\ge\frac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

\(\Leftrightarrow3\ge\frac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

\(\Leftrightarrow3\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\left(\sqrt[3]{abc}+1\right)\)

\(\Leftrightarrow\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\sqrt[3]{abc}+1\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\left(\sqrt[3]{abc}+1\right)^3\)

Dấu "=" xảy ra <=> a = b = c

Ko đăng câu hỏi linh tinh !

#Minh#

thi đấu không