Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2+b^2}{ab\left(a+b\right)^3}\ge\dfrac{2ab}{ab\left(a+b\right)^3}=\dfrac{2}{\left(a+b\right)^3}\\\dfrac{b^2+c^2}{bc\left(b+c\right)^3}\ge\dfrac{2bc}{bc\left(b+c\right)^3}=\dfrac{2}{\left(b+c\right)^3}\\\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\ge\dfrac{2ca}{ca\left(c+a\right)^3}=\dfrac{2}{\left(c+a\right)^3}\end{matrix}\right.\)
\(\Rightarrow VT\ge2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\)
Chứng minh rằng \(2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{9}{8}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{{}\begin{matrix}2ab\le a^2+b^2\\2bc\le b^2+c^2\\2ca\le c^2+a^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab\le a^2-ab+b^2\\bc\le b^2-bc+c^2\\ca\le c^2-ca+a^2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ab\left(a+b\right)\le\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\\bc\left(b+c\right)\le\left(b+c\right)\left(b^2-bc+c^2\right)=b^3+c^3\\ca\left(c+a\right)\le\left(c+a\right)\left(c^2-ca+a^2\right)=c^3+a^3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3ab\left(a+b\right)\le3\left(a^3+b^3\right)\\3bc\left(b+c\right)\le3\left(b^3+c^3\right)\\3ca\left(c+a\right)\le3\left(c^3+a^3\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^3+3ab\left(a+b\right)+b^3\le4\left(a^3+b^3\right)\\b^3+3bc\left(b+c\right)+c^3\le4\left(b^3+c^3\right)\\c^3+3ca\left(c+a\right)+a^3\le4\left(c^3+a^3\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^3\le4\left(a^3+b^3\right)\\\left(b+c\right)^3\le4\left(b^3+c^3\right)\\\left(c+a\right)^3\le4\left(c^3+a^3\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left(a+b\right)^3}\ge\dfrac{1}{4\left(a^3+b^3\right)}\\\dfrac{1}{\left(b+c\right)^3}\ge\dfrac{1}{4\left(b^3+c^3\right)}\\\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{1}{4\left(c^3+a^3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\)
Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\ge\dfrac{9}{8}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\ge\dfrac{9}{2\left(a^3+b^3+c^3\right)}=\dfrac{9}{2}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\ge\dfrac{9}{8}\) ( đpcm )
Vậy \(2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\ge\dfrac{9}{4}\)
Mà \(VT\ge2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\)
\(\Rightarrow VT\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{ab\left(a+b\right)^3}+\dfrac{b^2+c^2}{bc\left(b+c\right)^3}+\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\ge\dfrac{9}{4}\) ( đpcm )
