Question

[nguyễn thị ngọc trâm]

Cho ba số thực dương a,b,c thõa 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 Max của biểu thức:

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

Gỉai giúp mình nha

Answer 1

[Mr Lazy]

\(3\left(2a^2+b^2\right)=\left(1^2+1^2+1^2\right)\left(a^2+a^2+b^2\right)\ge\left(a+a+b\right)^2=\left(2a+b\right)^2\)

\(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

\(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)

\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)

\(gt\rightarrow7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\)

\(\Leftrightarrow7\left(x+y+z\right)^2=20\left(xy+yz+zx\right)+2015\)

Ta có: \(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\)

\(\Rightarrow7\left(x+y+z\right)^2\le\frac{20}{3}\left(x+y+z\right)^2+2015\)

\(\Leftrightarrow\frac{1}{3}\left(x+y+z\right)^2\le2015\)

\(\Leftrightarrow x+y+z\le\sqrt{6045}\)

\(P\le\frac{1}{3}\left(x+y+z\right)\le\frac{\sqrt{6045}}{3}\)

Dấu bằng xảy ra khi \(x=y=z=\frac{\sqrt{6045}}{3}\)hay \(a=b=c=\left(\frac{\sqrt{6045}}{3}\right)^{-1}\)