Áp dụng AM - GM

\(a+b+c\ge3\sqrt[3]{abc}\)

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

\(\Rightarrow P\ge9\)

(Dấu "="\(\Leftrightarrow a=b=c\))

Phá ngoặc ra ông giáo ạ:3

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

\(=3+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)

\(\ge3+3\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\) ( hồn nhiên cô si )

\(\ge3+3\sqrt[3]{\frac{8abc}{abc}}=9\) ( hồn nhiên cô si tiếp )

Dấu "=" xảy ra tại a=b=c

Ta có

\(P=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

mặt khác \(\frac{x}{y}+\frac{y}{x}\ge2\) với mọi x,y dương \(\Rightarrow\frac{P}{3+2+2+2}=9\)

Vậy P_min=9 khi và chỉ khi a=b=c

a Vi \(c^{2008}\) chua so mu chan \(\Rightarrow c^{2008}>0\Rightarrow a.b>0\) \(\Rightarrow\) a va b la 2 so nguyen cung dau \(\Rightarrow\) a va b la 2 so nguyen am \(\Rightarrow\) c la so nguyen duong

Vay a;b la so nguyen am;c la so nguyen duong

b,Vi |a|>0\(\Rightarrow\text{|a|}^{2009}>0\Rightarrow b.c>0\Rightarrow\)b va c la 2 so nguyen cung sau \(\Rightarrow\) b va c la 2 so nguyen am \(\Rightarrow\) a la so nguyen duong

Vay b;c la 2 so nguyen am;a la so nguyen duong

Nho tick cho minh nha

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Áp dụng AM-GM có:

\(2a^2+2b^2\ge4ab\)

\(8b^2+\dfrac{1}{2}c^2\ge4bc\)

\(8a^2+\dfrac{1}{2}c^2\ge4ac\)

Cộng vế với vế \(\Rightarrow VT\ge4\left(ab+bc+ac\right)=4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}ab+bc+ac=1\\a=b=\dfrac{c}{4}\end{matrix}\right.\)\(\Rightarrow a=b=\dfrac{1}{3};c=\dfrac{4}{3}\)