\(\left\{{}\begin{matrix}a^2=x\\b^2=y\\c^2=z\end{matrix}\right.\Rightarrow VT=\dfrac{x^3}{y+z}+\dfrac{y^3}{x+z}+\dfrac{z^3}{x+y}=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{xy+yz}+\dfrac{z^4}{xz+yz}\)
\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{xy+yz+zx}{2}\)
\(xy+yz+zx=a^2b^2+b^2c^2+c^2a^2\)
Ta can cm \(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
That vay
\(a^2b^2+b^2c^2\ge2b^2ac\)
\(b^2c^2+c^2a^2\ge2c^2ab\)
\(a^2b^2+a^2c^2\ge2a^2bc\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\left(dpcm\right)\)
\(\Rightarrow\sum\dfrac{a^6}{b^2+c^2}\ge\dfrac{abc\left(a+b+c\right)}{2}\)
Đúng 2
Bình luận (0)