SOS helps ^^
\(\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{a+c}+\dfrac{c^2+a^2}{a+b}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}-b+\dfrac{b^2+c^2}{a+c}-c+\dfrac{c^2+a^2}{a+b}-a\ge0\)
\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a+c\right)-\left(c-a\right)\left(a+b\right)}{b+c}\ge0\)
\(\Leftrightarrow\sum\left(a-b\right)\left(\dfrac{a+c}{b+c}-\dfrac{b+c}{a+c}\right)\ge0\)
\(\Leftrightarrow\sum\left(a-b\right)^2\dfrac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge0\)
Cần thêm \(a;b;c\) dương nha
\(\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}\ge\dfrac{\left(a+b+b+c+c+a\right)^2}{4\left(a+b+c\right)}=a+b+c\)
Thêm đk: a, b, c > 0.
Ta có: \(VT=\frac{1}{2}\left(\Sigma_{cyc}\frac{\left(a+b\right)^2}{b+c}+\Sigma_{cyc}\frac{\left(a-b\right)^2}{b+c}\right)\)
\(\ge\frac{1}{2}\left[\frac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)}+\frac{\left(a-b+b-c+a-c\right)^2}{2\left(a+b+c\right)}\right]\)
\(=\frac{\left(a+b+c\right)^2+\left(a-c\right)^2}{\left(a+b+c\right)}\). Vậy ta chứng minh: \(\frac{\left(a+b+c\right)^2+\left(a-c\right)^2}{\left(a+b+c\right)}\ge\left(a+b+c\right)\)
\(\Leftrightarrow\left(a-c\right)^2\ge0\) which is obvious!
Equality holds when \(a=b=c\)
q.e.d./.
