Ta có: 

Theo bất đẳng thức Cô - si, ta có: \(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\le\frac{a+b+a+c}{2}+\frac{b+c}{2}=1\)

\(\Rightarrow\sqrt{a}\left(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\right)\le\sqrt{a}\)hay \(\sqrt{a^2+abc}+\sqrt{abc}\le\sqrt{a}\)

Tương tự ta có: \(\sqrt{b^2+abc}+\sqrt{abc}\le\sqrt{b}\);\(\sqrt{c^2+abc}+\sqrt{abc}\le\sqrt{c}\)

Mà \(abc\le\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}\Rightarrow\sqrt{abc}\le\frac{1}{3\sqrt{3}}\)

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\le3\left(a+b+c\right)=3\)\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)

a=b=c=1/3

tôi đúng mừ

Ta có: (a+b)(a+c)=a²+ac+ab+bc=a(a+b+c)+bc=1/bc+bc

Mà b,c >0 nên 1/bc+bc>=2(tổng hai số nghịch đảo)

=> P>=2

Vậy GTNN của P=2

\(\dfrac{1}{1+a}=1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\)

Tương tự:

\(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ac}{\left(1+a\right)\left(1+c\right)}}\) ; \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+c\right)}}\)

Nhân vế với vế:

\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Rightarrow abc\le\dfrac{1}{8}\)

\(N_{max}=\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{2}\)

\(P=\dfrac{1}{abc}+\dfrac{1}{a^2+b^2+c^2}=\dfrac{a+b+c}{abc}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\left(1\right)\)

\(\)\(\left\{{}\begin{matrix}a+b+c=1\\\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow P\ge\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{2\left(ab+bc+ac\right)}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)

\(\Rightarrow P\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)

\(\Rightarrow P\ge9+\dfrac{17}{2\left(ab+bc+ac\right)}\)

mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)

\(\Rightarrow P\ge9+\dfrac{17}{2.\dfrac{1}{3}}=9+\dfrac{17.3}{2}=\dfrac{18+17.3}{2}=\dfrac{69}{2}\)

\(\Rightarrow Min\left(P\right)=\dfrac{69}{2}\)

\(a^{2012}+b^{2012}+c^{2012}\ge3\sqrt[3]{\left(abc\right)^{2012}}=3\)

\(\Rightarrow\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\le\dfrac{1}{3}\)

\(\Rightarrow-\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge-\dfrac{1}{3}\)

Lại có:

\(a^{2013}+a^{2013}+...+a^{2013}\left(\text{2012 số hạng}\right)+1\ge2013\sqrt[2013]{\left(a^{2013}\right)^{2012}}=2013.a^{2012}\)

\(\Rightarrow2012.a^{2013}+1\ge2013.a^{2012}\)

Tương tự: \(2012.b^{2013}+1\ge2013.b^{2012}\) ; \(2012.c^{2013}+1\ge2013.c^{2012}\)

Cộng vế với vế:

\(\Rightarrow a^{2013}+b^{2013}+c^{2013}\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012}\)

\(\Rightarrow A\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012\left(a^{2012}+b^{2012}+c^{2012}\right)}=\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{3}=1\)

\(A_{min}=1\) khi \(a=b=c=1\)

a=1

b=2

c=3

abc = ab + bc + ac 
Nếu a = b = c = 0 => thỏa mãn 

Nếu a, b, c khác 0 
=> Ta có: 
1 = (ab + bc + ca)/abc = 1/a + 1/b + 1/c 

Vậy {a; b; c} là tập hợp của {2; 3; 6}; {3; 3; 3}

sai roi le minh duc oi abc la a nhân b nhân c đó