Cho 3 số thực dương a,b,c thỏa mãn : \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015.\)
Tìm \(GTLN\) của biểu thức sau: \(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)
Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)
\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)
Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)
Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)
