@Nguyễn Thanh Hằng đọc xong xóa đii nha

Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3a^2b-3b^2a-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2-3ab\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\left(1\right)\)

C/m : \(a^2+b^2+c^2-ab-bc-ac\ge0\)

Giả sử điều phải c/m là đúng , ta có :

\(a^2+b^2+c^2-ab-bc-ac\ge0\)

\(\Rightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)\ge0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ( điều này luôn đúng )

\(\Rightarrow\) điều giả sử là đúng

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\ge0\left(2\right)\)

Từ ( 1 ) ; ( 2 )

\(\Rightarrow a+b+c=0\)

\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)

Lại có : \(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

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

\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)

\(=\dfrac{-abc}{abc}=-1\)

Vậy \(A=-1\)

ĐKXĐ: \(abc\ne0\)

\(a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

TH1: \(a+b+c=0\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)

TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

\(a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\left(a+b+c\ne0\right)\\ \Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\\ \Rightarrow a=b=c\\ \Rightarrow B=\dfrac{2}{a}.\dfrac{2}{b}.\dfrac{2}{c}=\dfrac{8}{abc}\)

Áp dụng BĐT AM - GM ta có:

$ \frac{a^3}{(1 + b)(1 + c)} + \frac{1 + b}{8} + \frac{1 + c}{8} \geq \frac{3}{4}a$

$\frac{b^3}{(1 + c)(1 + a)} + \frac{1 + c}{8} + \frac{1 + a}{8} \geq \frac{3}{4}b$

$\frac{c^3}{(1 + a)(1 + b)} + \frac{1 + a}{8} + \frac{1 + b}{8} \geq \frac{3}{4}c $

Cộng vế theo vế ta được:

$ P + \frac{2(a + b + c) + 6}{8} \geq \frac{3}{4}(a + b + c) $

$<=> P \geq \frac{1}{2}(a + b + c) - \frac{3}{4}$

$=> P \geq \frac{3}{4} (dpcm)$

\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2=3abc\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2ac-2bc=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\end{matrix}\right.\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

\(M=\dfrac{\left(a+b\right)}{b}.\dfrac{\left(b+c\right)}{c}.\dfrac{\left(a+c\right)}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{-abc}{abc}=-1\)

TH2: \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\Leftrightarrow a=b=c\)

\(M=\left(\dfrac{a}{a}+1\right)\left(\dfrac{a}{a}+1\right)\left(\dfrac{a}{a}+1\right)=2.2.2=8\)

Áp dụng bđt AM-GM cho 2 số dương:

\(a^3+b^3+c^3\ge3abc\)

Dấu "=" xảy ra khi:

\(a=b=c\)

Khi đó:

\(\left\{{}\begin{matrix}\dfrac{a}{b}=1\\\dfrac{b}{c}=1\\\dfrac{a}{c}=1\end{matrix}\right.\) \(\Leftrightarrow\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a+b+c=0\) hoặc \(a=b=c\) (bn tự chứng minh)

+) \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;a+c=-b\)\(\Rightarrow A=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)

\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)

+) \(a=b=c\Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)

Ta có \(\dfrac{1}{a^3\left(b+c\right)}=\dfrac{1}{\dfrac{1}{b^3c^3}\left(b+c\right)}=\dfrac{b^2c^2}{\dfrac{1}{b}+\dfrac{1}{c}}\)

Tương tự \(\Rightarrow VT=\dfrac{b^2c^2}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{c^2a^2}{\dfrac{1}{c}+\dfrac{1}{a}}+\dfrac{a^2b^2}{\dfrac{1}{a}+\dfrac{1}{b}}\)

\(\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}\) (BĐT B.C.S)

\(=\dfrac{\left(ab+bc+ca\right)^2}{2\left(\dfrac{ab+bc+ca}{abc}\right)}\)

\(=\dfrac{ab+bc+ca}{2}\) (do \(abc=1\))

\(\ge\dfrac{3\sqrt[3]{abbcca}}{2}\)

\(=\dfrac{3\left(\sqrt[3]{abc}\right)^2}{2}=\dfrac{3}{2}\) (do \(abc=1\))

ĐTXR \(\Leftrightarrow a=b=c=1\)