Question

Cho a,b,c > 0 thỏa mãn \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=3\)

Tìm GTLN của biểu thức: P = \(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\)

Answer 1

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

\(P\le\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{2}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(P\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{4}\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{3}{4}\)

\(P_{max}=\frac{3}{4}\) khi \(a=b=c=1\)