$(*)$Cho $a,b,c>0, a+b+c=2$. Chứng minh: 

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

Không mất tính tổng quát giả sử $a\ge b\ge c>0$

$\to \dfrac{1}{b+c}\ge \dfrac{1}{a+c}\ge \dfrac{1}{a+b}\to \dfrac{a}{b+c}\ge \dfrac{b}{a+c}\ge \dfrac{c}{a+b}$

Áp dụng BĐT Chev ta được:

$a.\dfrac{a}{b+c}+b.\dfrac{b}{a+c}+c.\dfrac{c}{a+b}\ge \dfrac{1}{3}(a+b+c)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)$

Áp dụng BĐT Chev ta được:

$a.\dfrac{1}{b+c}+b.\dfrac{1}{a+c}+c.\dfrac{1}{a+b}\ge \dfrac{1}{3}(a+b+c)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)$

Áp dụng BĐT Cauchy Schwars ta được:

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge \dfrac{1}{3}.(a+b+c).\dfrac{9}{2(a+b+c)}=\dfrac{3}{2}\\\to \dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge \dfrac{1}{3}.2.\dfrac{3}{2}=1$

Dấu "=" xảy ra khi: $a=b=c$

a,

Không mất tính tổng quát giả sử $a\ge b\ge c>0$

$\to \dfrac{1}{b+c}\ge \dfrac{1}{a+c}\ge \dfrac{1}{a+b}\to \dfrac{a}{b+c}\ge \dfrac{b}{a+c}\ge \dfrac{c}{a+b}$

Áp dụng BĐT Chev ta được:

$a.\dfrac{a}{b+c}+b.\dfrac{b}{a+c}+c.\dfrac{c}{a+b}\ge \dfrac{1}{3}(a+b+c)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)=\dfrac{2}{3}\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)$

Áp dụng BĐT Cauchy Schwars ta được:

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge \dfrac{1}{3}.(a+b+c).\dfrac{9}{2(a+b+c)}=\dfrac{3}{2}\\\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge \dfrac{1}{3}.(a+b+c).\dfrac{3}{2}=\dfrac{a+b+c}{2}$

Dấu "=" xảy ra khi: $a=b=c$