Kurosaki Akatsu giải thế thì đề bài cho \(b^2+c^2\le a^2\) để làm gì?
Áp dụng bất đẳng thức AM-GM ta có :
\(P=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(P=\frac{b^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{a^2}{c^2}\ge4.\sqrt[4]{\frac{b^2}{a^2}.\frac{c^2}{a^2}.\frac{a^2}{b^2}.\frac{a^2}{c^2}}=4.1=4\)
=> \(Min_P=4\)
Với a, b, c thực dương áp dụng BĐT Cô-si ta có:
\(P=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)=\left(\frac{b^2}{a^2}+\frac{c^2}{a^2}\right)+\left(\frac{a^2}{b^2}+\frac{a^2}{c^2}\right)\)
\(\ge2\sqrt{\frac{b^2}{a^2}.\frac{c^2}{a^2}}+2\sqrt{\frac{a^2}{b^2}.\frac{a^2}{c^2}}=2\left(\frac{bc}{a^2}+\frac{a^2}{bc}\right)\)
\(=2\left[\left(\frac{bc}{a^2}+\frac{a^2}{4bc}\right)+\frac{3a^2}{4bc}\right]\ge2\left(2.\sqrt{\frac{bc}{a^2}.\frac{a^2}{4bc}}+\frac{3\left(b^2+c^2\right)}{4bc}\right)\) (vì \(a^2\ge b^2+c^2\))
\(=2\left(2\sqrt{\frac{1}{4}}+\frac{3.2bc}{4bc}\right)\) (vì \(b^2+c^2\ge2bc\))
\(=2\left(2.\frac{1}{2}+\frac{3}{2}\right)=5\)
Vậy Pmin = 5
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}a^2=b^2+c^2\\b=c\end{cases}}\)
