Question

cho a,b,c >0,a+b+c=3. tìm min Q= \(\dfrac{a^5}{b+c}+\dfrac{b^5}{c+a}+\dfrac{c^5}{a+b}\)

 

Answer 1

Ta có: \(a^3+1+1\ge3a\) ; tương tự: \(b^3+2\ge3b\) ; \(c^3+2\ge3c\)

\(\Rightarrow a^3+b^3+c^3\ge3\left(a+b+c\right)-6=3\)

\(Q=\dfrac{a^6}{ab+ac}+\dfrac{b^6}{bc+ab}+\dfrac{c^6}{ac+bc}\ge\dfrac{\left(a^3+b^3+c^3\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3^2}{\dfrac{2}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)