Áp dụng t/c dtsbn:

\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\Rightarrow a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a\cdot a\cdot a}=\dfrac{8a^3}{a^3}=8\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)

Áp dụng tính chất dãy tỉ số bằng nhau:

(a + b - c)/c = (a + c - b)/b = (b + c - a)/a = (a + b - c + a + c - b + b + c - a)/(a + b + c) = 1

--> a + b - c = c

a + c - b = b

b + c - a = a

--> a + b = 2c

a + c = 2b

b + c = 2a

Ta có: P = (a + b)(b + c)(a + c)/(abc) = 2c.2a.2b/(abc) = 8

Có: \(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)

Áp dụng tính chất của dãy tỉ số bằng nhau , ta được:

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

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

Xét: a + b + c = 0 \(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)(1)

Thay (1) vào A, ta có:

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

Xét a + b + c ≠ 0: 

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

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

\(\Rightarrow\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)(2)

Thay (2) vào A, ta có:

\(A=\dfrac{2c.2a.2b}{abc}=8\)

Vậy...

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

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

Do đó: 

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

Thay a+b=2c;b+c=2a và c+a=2b vào biểu thức \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+b\right)}{abc}\), ta được: 

\(P=\dfrac{2a\cdot2b\cdot2c}{abc}=\dfrac{8abc}{abc}=8\)

Vậy: P=8

Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) = \(\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\) (t/c dãy tỉ số bằng nhau)

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

Ta cũng có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+b+c-a}{a+c}\) (t/c dãy tỉ số bằng nhau)

hay \(\dfrac{a+b-c}{c}=\dfrac{2b}{a+c}\) (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{2b}{a+c}=1\) \(\Leftrightarrow\) a + c = 2b (*)

Tương tự ta cũng có: a + b = 2c (**); b + c = 2a (***)

Thay (*); (**); (***) vào P ta được:

P = \(\dfrac{2a.2b.2c}{abc}\) = 2.2.2 = 8

Vậy P = 8

Chúc bn học tốt!

TH1: \(a+b+c=0\Rightarrow P=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

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

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\) 

\(\Rightarrow P=\dfrac{2a.2b.2c}{abc}=8\)

Lời giải:

Áp dụng BĐT AM-GM ta có:

$\text{VT}=[\frac{a+1}{4}+\frac{1}{a+1}+\frac{3}{4}a-\frac{1}{4}][\frac{b+1}{4}+\frac{1}{b+1}+\frac{3}{4}b-\frac{1}{4}][\frac{c+1}{4}+\frac{1}{c+1}+\frac{3}{4}c-\frac{1}{4}]$

$\geq [2\sqrt{\frac{1}{4}}+\frac{3}{4}a-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}b-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}c-\frac{1}{4}]$
$=\frac{3}{4}(a+1).\frac{3}{4}(b+1).\frac{3}{4}(c+1)$
$=\frac{27}{64}(a+1)(b+1)(c+1)$

$\geq \frac{27}{64}.2\sqrt{a}.2\sqrt{b}.2\sqrt{c}$

$=\frac{27}{64}.8\sqrt{abc}\geq \frac{27}{64}.8=\frac{27}{8}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$

\(a+\dfrac{1}{a+1}=\dfrac{a^2+a+1}{a+1}=\dfrac{4a^2+4a+4}{4\left(a+1\right)}=\dfrac{3\left(a+1\right)^2+\left(a-1\right)^2}{4\left(a+1\right)}\ge\dfrac{3\left(a+1\right)^2}{4\left(a+1\right)}=\dfrac{3}{4}\left(a+1\right)\ge\dfrac{3}{2}\sqrt{a}\)

Tương tự: \(b+\dfrac{1}{b+1}\ge\dfrac{3}{2}\sqrt{b}\) ; \(c+\dfrac{1}{c+1}\ge\dfrac{3}{2}\sqrt{c}\)

Nhân vế:

\(VT\ge\dfrac{27}{8}\sqrt{abc}\ge\dfrac{27}{8}\) (đpcm)

\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)

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

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

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

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

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

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

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

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

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

\(=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)

Trường hợp 1: a+b+c=0

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

\(\Leftrightarrow N=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}=-1\)

Trường hợp 2: a=b=c

\(\Leftrightarrow N=\dfrac{b+b}{b}\cdot\dfrac{a+a}{a}\cdot\dfrac{c+c}{c}=2\cdot2\cdot2=8\)

1, Ta có a^3+b^3+c^3=3abc

-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2

-> (a+b)^{3 +} c^3 - 3ab(a+b+c)=0

-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0

-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0

Th1: a+b+c=0

->P= a+b/2 . b+c/2 . c+a/2

= (-c)(-a)(-b)/2=-1

TH2 a^2+b^2+c^2-ab-bc-ca=0

->2a^2+2b^2+2c^2-2ab-abc-2ac=0

->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0

-> (a-b)^2+(a-c)^2+(b-c)^2=0

Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0

Dấu = xảy ra (=)a-b=0

                         b-c=0

                          a-c=0

-> a=b=c

->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8