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

\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)

Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)

Biến đổi tương đương:

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

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

\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(a=b=c\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{ab+ac+bc}\ge3\)

b/ \(VT=\frac{\left(a+b+c\right)^2}{ab+ac+bc}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}=\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}\)

\(\Rightarrow VT\ge\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+2\sqrt{\frac{\left(a+b+c\right)^2\left(ab+ac+bc\right)}{9\left(ab+ac+bc\right)\left(a+b+c\right)^2}}\ge\frac{8.3}{9}+\frac{2}{3}=\frac{10}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)

Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.

Lời giải vắn tắt:

\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)

( thay \(a^2+b^2+c^2=1\))

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

<=> a^2 + b^2 + c^2 >= ab + bc + ac

<=> 2a^2 + 2b^2 + 2c^2 >= 2ab + 2bc + 2ac

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

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

( luôn đúng với mọi a ; b ; c )

( đpcm )

2 ) P =  \(\frac{\left(a+b+c\right)^2}{ab+bc+ac}+\frac{ab+bc+ac}{\left(a+b+c\right)^2}=\frac{\left(a+b+c\right)^2}{9\left(ab+bc+ac\right)}+\frac{ab+bc+ac}{\left(a+b+c\right)^2}+\frac{8\left(a+b+c\right)^2}{9\left(ab+bc+ac\right)}\)

AD BĐT Cô - si và BĐT phụ đã cmt ở trên  ta có : \(P\ge2.\frac{1}{3}+\frac{8.3.\left(ab+bc+ac\right)}{9\left(ab+bc+ac\right)}=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)

Dấu " = " xảy ra <=> a = b = c 

Khôi Bùi : theo e ý 2 có thể đơn giản hóa vấn đề bằng cách đặt ẩn phụ

đặt \(\frac{\left(a+b+c\right)^2}{ab+bc+ca}=t\left(t\ge3\right)\)

\(\Rightarrow P=t+\frac{1}{t}=\frac{t}{9}+\frac{1}{t}+\frac{8}{9}t\)

Áp dụng BĐT AM-GM ta có:

\(P\ge2.\sqrt{\frac{t}{9}.\frac{1}{t}}+\frac{8}{9}t\ge\frac{2.1}{3}+\frac{8}{9}.3=\frac{10}{3}\)

Dấu " = " xảy ra <=> a=b

Ta có a(a² - bc) + b(b² - ca) + c(c² - ab) 

= a³ + b³ + c³ - 3abc

= (a + b)³ - 3ab(a + b) + c³ - 3abc

= [(a + b)³ + c³] - 3ab(a + b + c) 

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

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

= (a + b + c)(a² +  b² + c² - ab - ac - bc)  (đpcm)