Question

Cho a, b, c là các số dương thỏa mãn \(b^2+c^2\le a^2\). Tìm GTNN của biểu thức: \(P=\dfrac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)

Answer 1

giúp mình theo BĐT Côsi nhá 

Answer 2

\(b^2+c^2\le a^2\Leftrightarrow\dfrac{b^2}{a^2}+\dfrac{c^2}{a^2}\le1\)

Đặt \(\left(\dfrac{b^2}{a^2};\dfrac{c^2}{a^2}\right)=\left(m;n\right)\Leftrightarrow m+n\le1\)

\(\Leftrightarrow P=m+n+\dfrac{1}{m}+\dfrac{1}{n}=\left(m+\dfrac{1}{4m}\right)+\left(n+\dfrac{1}{4n}\right)+\dfrac{3}{4}\left(\dfrac{1}{m}+\dfrac{1}{n}\right)\\ \Leftrightarrow P\ge2\sqrt{m\cdot\dfrac{1}{4m}}+2\sqrt{n\cdot\dfrac{1}{4n}}+\dfrac{3}{4}\cdot\dfrac{4}{m+n}\\ \Leftrightarrow P\ge2\cdot\dfrac{1}{2}+2\cdot\dfrac{1}{2}+\dfrac{3}{4}\cdot4=5\)

Dấu \("="\Leftrightarrow m=n=\dfrac{1}{2}\Leftrightarrow b=c=\sqrt{2}a\)