Với mọi a;b;c dương ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
Đồng thời: \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow\left(a+b+c\right)^3\ge27abc\Rightarrow\dfrac{1}{abc}\ge\dfrac{27}{\left(a+b+c\right)^3}\)
Do đó:
\(VT=\dfrac{a^2+b^2+c^2}{2}+\dfrac{a^2+b^2+c^2}{abc}\ge\dfrac{\left(a+b+c\right)^2}{6}+\dfrac{\left(a+b+c\right)^2}{3abc}\ge\dfrac{\left(a+b+c\right)^2}{6}+\dfrac{9\left(a+b+c\right)^2}{\left(a+b+c\right)^3}\)
\(VT\ge\dfrac{\left(a+b+c\right)^2}{6}+\dfrac{9}{a+b+c}=\dfrac{\left(a+b+c\right)^2}{6}+\dfrac{9}{2\left(a+b+c\right)}+\dfrac{9}{2\left(a+b+c\right)}\)
\(VT\ge3\sqrt[3]{\dfrac{81\left(a+b+c\right)^2}{24\left(a+b+c\right)^2}}=\dfrac{9}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
a,b,c > 0 . C/m đề nha mng nãy quên ghi đề:>>
Đề bài thiếu, BĐT này chỉ đúng với a;b;c dương
