Question

[Đỗ Thị Kim Tiên]

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

Cm \(\frac{1}{2a^2+b^2}\)+\(\frac{1}{2b^2+c^2}\)+\(\frac{1}{2c^2+a^2}\)<=\(\frac{1}{9}\)

2. Tìm nghiệm nguyên của phương trình

x⁴+2x³+2x²+x+3=y²

Answer 1

[Đinh quang hiệp]

\(\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}=\frac{1}{a^2+a^2+b^2}+\frac{1}{b^2+b^2+c^2}+\frac{1}{c^2+c^2+a^2}\)

\(< =\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{9}\left(\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{1}{9}\left(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\)(bđt svacxo)

\(=\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)=\frac{1}{9}\cdot3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

\(=\frac{1}{9}\cdot3\cdot\frac{1}{3}=\frac{1}{9}\cdot1=\frac{1}{9}\)

\(\Rightarrow\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}< =\frac{1}{9}\)(đpcm)

dấu = xảy ra khi \(\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{1}{9}\Rightarrow a=b=c=3\)