Do vai trò a;b;c là như nhau, ko mất tính tổng quát giả sử \(c=min\left\{a;b;c\right\}\)
Đặt \(\left\{{}\begin{matrix}a+\dfrac{c}{2}=x>0\\b+\dfrac{c}{2}=y>0\end{matrix}\right.\)
\(\Rightarrow a+b+c=x+y\)
\(x^2=\left(a+\dfrac{c}{2}\right)^2=a^2+ac+\dfrac{c^2}{4}\ge a^2+c^2+\dfrac{c^2}{4}\ge a^2+c^2\) (do \(a\ge c\))
\(\Rightarrow\dfrac{1}{a^2+c^2}\ge\dfrac{1}{x^2}\)
\(y^2=\left(b+\dfrac{c}{2}\right)^2=b^2+bc+\dfrac{c^2}{4}\ge b^2+c^2+\dfrac{c^2}{4}\ge b^2+c^2\) (do \(b\ge c\))
\(\Rightarrow\dfrac{1}{b^2+c^2}\ge\dfrac{1}{y^2}\)
\(x^2+y^2=\left(a+\dfrac{c}{2}\right)^2+\left(b+\dfrac{c}{2}\right)^2=a^2+b^2+c\left(a+b\right)+\dfrac{c^2}{2}\ge a^2+b^2\)
\(\Rightarrow\dfrac{1}{a^2+b^2}\ge\dfrac{1}{x^2+y^2}\)
\(\Rightarrow\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\ge\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\) (1)
Lại có:
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}=\dfrac{3}{4}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{1}{4}\left(\dfrac{x^2+y^2}{x^2y^2}+\dfrac{4}{x^2+y^2}\right)\)
\(\ge\dfrac{3}{4}.2\sqrt{\dfrac{1}{x^2y^2}}+\dfrac{1}{4}.2\sqrt{\dfrac{4\left(x^2+y^2\right)}{x^2y^2.\left(x^2+y^2\right)}}=\dfrac{5}{2xy}\ge\dfrac{10}{\left(x+y\right)^2}=\dfrac{10}{\left(a+b+c\right)^2}\) (2)
(1);(2) \(\Rightarrow\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\ge\dfrac{10}{\left(a+b+c\right)^2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b>0;c=0\) và các hoán vị của chúng
