Question

[dbrby]

cho a,b,c >0 thỏa mãn \(b^2+c^2\le a^2\)

Tìm GTNN của \(P=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2\le1\)

Đặt \(\left[\left(\frac{b}{a}\right)^2;\left(\frac{c}{a}\right)^2\right]=\left(x;y\right)\Rightarrow x+y\le1\)

\(P=x+y+\frac{1}{y}+\frac{1}{x}\ge x+y+\frac{4}{x+y}\)

\(P\ge x+y+\frac{1}{x+y}+\frac{3}{x+y}\ge2\sqrt{\frac{x+y}{x+y}}+\frac{3}{1}=5\)

\(p_{min}=5\) khi \(x=y=\frac{1}{2}\Leftrightarrow b=c=\frac{a}{\sqrt{2}}\)