TK

\(\Leftrightarrow a^3+b^3+c^3+6abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)

\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)

Đây là BĐT Schur bậc 3, cách chứng minh nó có thể tìm thấy ở mọi nơi

Vì a = b+c => b = a-c

Ta có : c = bd/ b-d

=>c/d = b/b-d

=> c/d = a-c / b-d = c +a-c / d +b-d = a/b

Vậy a/b = c/d

Nhớ like cho mình

điều kiên:
b<>d <>0
=> c<>0
a=b+c
=> a<>0
*
c=(b.d):(b-d).
=> c*(b-d)=b*d
=>cb-cd=b*d
=>cb=cd+bd
=>=cb=d(b+c)=ad (vì b+c=a)
cb=ad (từ cái này xoay kiểu gì cũng được)
c:d=a:b
a/b=c/d >>>dpcm
c/a=d/b

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

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

\(\dfrac{1}{a^2+b^2-c^2}+\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}\)

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

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

\(=\dfrac{1}{a^2+b^2-a^2-2ab-b^2}+\dfrac{1}{b^2+c^2-b^2-2bc-c^2}+\dfrac{1}{c^2+a^2-c^2-2ac-a^2}\)

\(=\dfrac{1}{-2ab}+\dfrac{1}{-2bc}+\dfrac{1}{-2ac}\)

\(=\dfrac{c+a+b}{-2abc}=\dfrac{0}{-2abc}=0\)

ta có a+b+c=0=>a+b=-c =>(a+b)^2=c^2=> a^2+b^2=c^2-2ab =>a^2+b^2-c^2=-2ab
tương tự ta sẽ có

-1/2ab-1/2bc-1/2ac =-c/2abc- a/2abc- b/2abc =0 (vì a+b+c=0)