Cách 1:
Áp dụng bđt Bunhiacopxki :
\(VT=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\cdot\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Cách 2:
Áp dụng bđt Cô-si :
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\cdot\left(b+c\right)}{4\cdot\left(b+c\right)}}=a\)
Tương tự : \(\frac{b^2}{c+a}+\frac{c+a}{4}\ge b\); \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
Cộng vế :
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge a+b+c-\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Cách 1: Svac:
\(VT\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Đẳng thức xảy ra khi a = b = c
Cách 2: SOS:
\(VT-VP=\left(\frac{a^2}{b+c}-\frac{a}{2}\right)+\left(\frac{b^2}{c+a}-\frac{b}{2}\right)+\left(\frac{c^2}{a+b}-\frac{c}{2}\right)\)
\(=\Sigma_{cyc}\left(\frac{a\left(a-b\right)}{2\left(b+c\right)}-\frac{b\left(a-b\right)}{2\left(c+a\right)}\right)=\Sigma\frac{\left(a-b\right)^2\left(a+b+c\right)}{2\left(b+c\right)\left(c+a\right)}\ge0\)
Vậy có đpcm.
Cách 3: Đợi tí em show hàng phương pháp mới:D
Giả sử \(c=min\left\{a,b,c\right\}\)
\(VT-VP=\frac{\left(a-b\right)^2\left(a+b+c\right)\left(7a+7b-2c\right)+\left(a+b-2c\right)^2\left(a+b+c\right)\left(a+b+2c\right)}{8\left(a+b\right)\left(a+c\right)\left(b+c\right)}\ge0\)
