Bài làm

Ta có : a³ + b³ + c³ = 3abc

<=> ( a³ + b³ ) + c³ - 3abc = 0

<=> ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

<=> [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

<=> ( a + b + c )[ ( a + b )² - ( a + b )c + c² ] - 3ab( a + b + c ) = 0

<=> ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

<=> ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Vì a, b, c dương => a + b + c > 0 => a + b + c = 0 vô lí

Xét a² + b² + c² - ab - bc - ac = 0

<=> 2( a² + b² + c² - ab - bc - ac ) = 2.0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

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

VT ≥ 0 ∀ a,b,c . Đẳng thức xảy ra <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Leftrightarrow a=b=c\)

=> \(P=\left(\frac{a}{b}-1\right)+\left(\frac{b}{c}-1\right)+\left(\frac{c}{a}-1\right)=\left(\frac{a}{a}-1\right)+\left(\frac{b}{b}-1\right)+\left(\frac{c}{c}-1\right)\)

\(=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)\)

\(=0\)

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

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

Nếu \(a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Rightarrow a=b=c\)

Khi đó \(A=2^3=8\)

Nếu \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

Thay vào ta được:

\(A=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-abc}{abc}=-1\)

Vậy A = 8 hoặc A = -1

Theo tính chất dãy tỉ số bằng nhau ta có : a+b-c/c = b+c-a/a = c+a-b/b = a+b-c+b+c-a+c+a-b/a+b+c = a+b+c/a+b+c = 1

Ta có : a+b-c/c=1  => a+b-c=c  => a+b+c=3c   (1)

Ta có : b+c-a/a=1  => b+c-a=a  => a+b+c=3a   (2)

Ta có : c+a-b/b=1  => c+a-b=b  => a+b+c=3b   (3)

Từ (1);(2);(3)   => 3c=3a=3b  => a=b=c  => b/a=1 ; a/c=1 ; c/b=1

=> B= (1+b/a)(1+a/c)(1+c/b)  = (1+1)(1+1)(1+1) = 2.2.2 = 8

=8

8 8 cái địt mẹ mày

Ồ sorry bạn nhiều, chỗ đấy bị lỗi kĩ thuật rồi, mình sửa lại nhé :

\(M\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)

Lại có : \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt{a^3b^3c^3}}{2}=\frac{3}{2}\)

Do đó : \(M\ge\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

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

Tương tự : \(\frac{1}{b^3\left(a+c\right)}=\frac{\left(\frac{1}{b}\right)^2}{b\left(a+c\right)}\) , \(\frac{1}{c^3\left(a+b\right)}=\frac{\left(\frac{1}{c}\right)^2}{c\left(a+b\right)}\)

Ta thấy : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Áp dụng BĐT Svacxo ta có :

\(M=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^2\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)   \(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Vâỵ \(M_{min}=\frac{3}{2}\) tại \(a=b=c=1\)

giải thích cho mình với, sao \(\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)  vậy