\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\) ; \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
Cộng vế với vế:
\(VT+\frac{a+b+c}{2}\ge a+b+c\Rightarrow VT\ge\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)