a/b-c + b/c-a + c/a-b=0 =>a/b-c=-(b/c-a + c/a-b)=c/a-b - b/c-a =b/a-c + c/b-a = b²-ab+ac-c²/(a-b)(c-a)

Tương tự rồi công lại

a/b-c+b/c-a+c/a-b=0

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

=c/a-b/c-a

=b/a-c+c/b-a

=b²-ab+ac-c²/(a-b) ( c - a )

Ta có:

\(\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\\=\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2\right]\left[\left(\frac{\sqrt{a}}{b+c}\right)^2+\left(\frac{\sqrt{b}}{b+c}\right)^2+\left(\frac{\sqrt{c}}{a+c}\right)^2\right]\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)

Mà ta có:

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\) (BĐT Nesbit)

\(\Rightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\ge\frac{9}{4}\\ \Rightarrow\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\ge\frac{9}{4}\)

\(\Rightarrow\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\ge\frac{9}{4\left(a+b+c\right)}\left(đpcm\right)\)

\(\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\)

\(=\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\left(\frac{a}{b+c}\right)^2+\frac{1}{4}+\left(\frac{b}{c+a}\right)^2+\frac{1}{4}+\left(\frac{c}{a+b}\right)^2+\frac{1}{4}+\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\frac{3}{4}\)

\(\ge2\cdot\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\frac{3}{4}=\frac{9}{4}\) ( dpcm )

Ad Lâm chuẩn hóa a + b + c = 3 rồi nên em chuẩn hóa a + b + c = 1 nha:D

Chuẩn hóa a + b + c = 1 \(\Rightarrow0< a,b,c< 1\)

BĐT quy về: \(\Sigma\frac{a}{\left(1-a\right)^2}\ge\frac{9}{4}\)

Ta chứng minh BĐT phụ sau:

\(\frac{a}{\left(1-a\right)^2}\ge\frac{9}{2}a-\frac{3}{4}\Leftrightarrow\frac{\left(27-18x\right)\left(x-\frac{1}{3}\right)^2}{4\left(1-x\right)^2}\ge0\)(đúng)

Thiết lập tương tự các BĐT còn lại và cộng theo vế thu được đpcm:)

Ta có

\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(a-c\right)\left(b-a\right)}\)

\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)

Tương tự

\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ab-a^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)

\(\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)

\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{b^2-ab+ac-c^2+c^2-bc+ab-a^2+a^2-ac+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(a-b\right)}\)

=0 ( ĐPCM)

Cái phân thức đầu tiên ở vế trái viết sai thì phải (ở cái tử phải là b²c chứ!).

Đặt A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)

     B = \(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

\(\Rightarrow\)A . B  = 9

Ta có : A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)

Nhân abc với A ta được:

 A_abc = \(\frac{abc\left(a-b\right)}{c}+\frac{abc\left(b-c\right)}{a}+\)\(\frac{abc\left(c-a\right)}{b}\)

A_abc =  ab.( a - b ) + bc.( b - c ) + ac.( c - a )

A_abc = ab.( a - b ) + bc.( a - c + b - a ) + ac.( a - c )

A_abc = ab.( a - b ) - bc.( a - b ) - bc.( c - a ) + ac.(c - a )

A_abc = b.( a - b ).( a - c ) - c.( a - b ).(c - a ) 

A_abc= ( a - b ).( a - c ).( b - c )

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

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

Nhân ( a - b ).( c - a ).( b - c ) với B ta được :

B( a - b).( c - a ).( b - c ) = \(\frac{c\left(a-b\right).\left(c-a\right).\left(b-c\right)}{a-b}\)+  \(\frac{a\left(a-b\right).\left(b-c\right).\left(c-a\right)}{b-c}\)+ \(\frac{b\left(a-b\right).\left(b-c\right).\left(c-a\right)}{c-a}\)

B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) + a.( b - c ).( c - a ) + b.( a - b ).( b - c)

B( a - b ).( c - a ) .( b - c ) = c.( c - a ).( b - c ) + ( a - b ).( -b - c ).( c - a ) + b.( a - b ).( b - c )

B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) - b.( a - b ).( c- a ) + b.( a - b ).(b - c ) - c.( a - b ).( c - a )

B( a - b ).( c - a ).( b - c ) = c.( c - a ).( -a + 2b - c ) + b.( a - 2c +b).(a - b )

B( a - b).( c - a ).( b - c ) = -3bc.( b + c - 2a )

B( a - b ).( c - a ).( b - c ) = -9abc

B = \(\frac{9abc}{\left(a-b\right).\left(c-a\right).\left(b-c\right)}\)

NHÂN A VỚI B :

\(\frac{\left(a-b\right).\left(b-c\right).\left(a-c\right)}{abc}\)\(.\)\(\frac{9abc}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)= 9

\(\Rightarrow\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right).\)\(\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)

MÌNH CŨNG KHÔNG CHẮC LẮM !