\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)

\(\Leftrightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=8\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=8abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ac+c^2\right)-8abc=0\)

\(\Leftrightarrow a^2b+abc+a^2c+ac^2+ab^2+b^2c+abc+bc^2-8abc=0\)

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

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

Do a;b;c dương nên \(b\left(a-c\right)^2;c\left(a-b\right)^2;a\left(b-c\right)^2\ge0\forall a;b;c\)

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

Đẳng thức xảy ra \(\Leftrightarrow a=b=c\) Thay vào P ta được :

\(P=\frac{a^3+a^3+a^3}{a.a.a}=\frac{3a^3}{a^3}=3\)

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

Do \(a,b,c\) là các số dương suy ra:

\(a>0;b>0;c>0\)

Suy ra: \(a+b+c>0\)

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

\(\Rightarrow a+b+c=0\) hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)

Do \(a+b+c>0\)

Suy ra: \(a^2+b^2+c^2-ab-bc-ca=0\)

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

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

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

Suy ra: \(a-b=0;b-c=0\) và \(c-a=0\)

Suy ra: \(a=b=c\)

Suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\)

Ta có: \(\left(\frac{a}{b}-1\right)+\left(\frac{b}{c}-1\right)+\left(\frac{c}{a}-1\right)=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)=0\)

Vậy ...

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 <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Leftrightarrow a=b=c\)

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

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

\(=0\)

Áp dụng BĐT AM-GM (Cô si): \(A\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(=3\sqrt[3]{\frac{1}{a\left(b+c\right).b\left(c+a\right).c\left(a+b\right)}}=\frac{3}{\sqrt[3]{\left(ab+ca\right)\left(bc+ab\right)\left(ca+bc\right)}}\)

\(\ge\frac{9}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

Đẳng thức xảy ra khi a = b = c = 1

P/s: Check giúp em xem có ngược dấu không:v

Cach khac 

Dat \(\left(ab;bc;ca\right)\rightarrow\left(x;y;z\right)\)

\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x^2+y^2+z^2\ge3\\xyz\le1\end{cases}}\)

Ta co:

\(A=\frac{1}{ab+b^2}+\frac{1}{bc+c^2}+\frac{1}{ca+a^2}\)

\(=\frac{1}{x+\frac{xy}{z}}+\frac{1}{y+\frac{yz}{x}}+\frac{1}{z+\frac{zx}{y}}\ge\frac{9}{3+xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\ge\frac{9}{3+3}=\frac{3}{2}\)

Dau '=' xay ra khi \(a=b=c=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(a=b=c=1\)

MaiLink bạn hãy chứng minh: \(xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le3\) xem thế nào? Nếu như ko c/m được thì bài này ngược dấu.