Question

chứng minh rằng với a,b,c>=0 thỏa mãn a+b+c=3.CMR: 4/(a+b)^3+4/(b+c)^3+4/(c+a)^3>=a/(b+c)+b/(c+a)+c/(a+b)

Answer 1

\(a;b;c>0\) \((a;b;c\ge0\) \(là\) \(sai)\)

\(\Sigma\dfrac{4}{\left(a+b\right)^3}\ge\Sigma\dfrac{c}{a+b}\Leftrightarrow\Sigma\dfrac{4}{\left(3-c\right)^3}\ge\Sigma\dfrac{c}{3-c}\left(1\right)\)

\(\dfrac{4}{\left(3-c\right)^3}\ge\dfrac{c}{3-c}\Leftrightarrow4\left(3-c\right)\ge c\left(3-c\right)^3\Leftrightarrow4\left(c-3\right)-c\left(3-c\right)^3\ge0\Leftrightarrow-\left(3-c\right)\left(c-4\right)\left(c-1\right)^2\ge0\left(2\right)\)

\(do:a,b,c>0;a+b+c=3\Rightarrow0< a,b,c< 3\Rightarrow\left(2\right)\) \(đúng\)

\(tương\) \(tự\Rightarrow\dfrac{4}{\left(3-a\right)^3}\ge\dfrac{a}{3-a};\dfrac{4}{\left(3-b\right)^3}\ge\dfrac{b}{3-b}\)

\(\Rightarrow\)\(\left(1\right)đúng\Rightarrowđpcm\)