Question

Cho a,b,c là 3 số thực dương thỏa mãn \(\dfrac{1}{a^2}\)+ \(\dfrac{1}{b^2}\) + \(\dfrac{1}{c^2}\)=1

Tìm GTLN : P = \(\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}\) + \(\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}\) + \(\dfrac{1}{\sqrt{5c^2+2ca+2a^2}}\)

Answer 1

Ta có :\(\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}=\dfrac{1}{\sqrt{\left(4a^2+4ab+b^2\right)+\left(a^2-2ab+b^2\right)}}\)

\(=\dfrac{1}{\sqrt{\left(2a+b\right)^2+\left(a-b\right)^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{2a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\) (Cosi)

Tương tự cộng lại ta được :

\(P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}=\dfrac{1}{\sqrt{3}}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\sqrt{3}\)