BĐT cần chứng minh tương đương:

\(a^2+b^2+c^2\ge2ab-2bc+2ca\)

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

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

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

Vậy BĐT đã cho đúng

Vì a,b,c là 3 cạnh tam giác nên \(a+b>c\Leftrightarrow ac+bc>c^2\)

CMTT: \(ab+bc>b^2;ab+ac>a^2\)

Cộng vế theo vế \(\Leftrightarrow a^2+b^2+c^2< ab+bc+ca+ab+bc+ca\)

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

Ta chứng minh 1 bđt phụ:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) (với a;b;c>0)
Thật vậy,ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\)

Mà: \(\frac{a}{b}+\frac{b}{a}\ge2;\frac{b}{c}+\frac{c}{b}\ge2;\frac{c}{a}+\frac{a}{c}\ge2\left(Cauchy\right)\)nên ta có đpcm 

Vậy bđt đc chứng minh
Áp dụng:

\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{a^2+b^2+c^2+2ab+2bc+2ac}=\frac{9}{\left(a+b+c\right)^2}\ge9\)

Dấu bằng khi a=b=c=1/3

a)

$A=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}$

$\ge \dfrac{(a+b+c)^2}{a(b^2+1)+b(c^2+1)+c(a^2+1)}\qquad (\text{Cauchy Engel})$

$=\dfrac{1}{ab^2+bc^2+ca^2+1}$

$\ge \dfrac{1}{ab(a+b)+bc(b+c)+ca(c+a)+1}$

$=\dfrac{1}{(a+b+c)(ab+bc+ca)+1}

$\ge \dfrac{1}{\frac13+1}$ $=\dfrac34$

Dấu bằng khi $a=b=c=\dfrac13$.

$\boxed{A_{\min}=\dfrac34}$

b)

$B=\dfrac{a}{ab+2c}+\dfrac{b}{bc+2a}+\dfrac{c}{ca+2b}$

$\ge \dfrac{(a+b+c)^2}{a(ab+2c)+b(bc+2a)+c(ca+2b)}$

$=\dfrac4{a^2b+b^2c+c^2a+2(ab+bc+ca)}$

Lại có $a^2b+b^2c+c^2a\le (a+b+c)(ab+bc+ca)$$=2(ab+bc+ca)$

Nên $B\ge \dfrac4{4(ab+bc+ca)}$$=\dfrac1{ab+bc+ca}$

$\ge \dfrac1{\frac{(a+b+c)^2}{3}}$ $=\dfrac34$

Dấu bằng khi $a=b=c=\dfrac23$.

$B_{\min}=\dfrac34$

Đặt \(m=a^2+bc\);\(n=b^2+2ca\);\(p=c^2+2ab\)

Lúc đó: \(m+n+p=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(=\left(a+b+c\right)^2< 1\)(vì a + b + c < 1 )

\(BĐT\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge9\)và m + n + p < 1 ; m,n,p > 0 

Áp dụng BĐT Cô -si cho 3 số không âm:

\(m+n+p\ge3\sqrt[3]{mnp}\)

và \(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge3\sqrt[3]{\frac{1}{mnp}}\)

\(\Rightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge9\)

Mà m + n + p < 1 nên \(\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge9\)

hay \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge9\)