Question

Cho a, b, c > 0; a + b + c = 3

Tìm GTNN ^{\(Q=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)}

Answer 1

\(Q=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Hoặc: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

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: \(Q+\frac{a+b+c}{2}\ge a+b+c\Rightarrow Q\ge\frac{a+b+c}{2}=\frac{3}{2}\)

 

Answer 2

rất tuyệt vời xin cảm ơn