Question

[Việt Khải Võ]

cho a khác 0; a+c>2; (2b^2-c^2)/a^2>=4. cmr a^2+b^2+c^2>4

Answer 1

[Nguyễn Việt Lâm]

\(\frac{2b^2-c^2}{a^2}\ge4\Leftrightarrow2b^2-c^2\ge4a^2\)

\(\Leftrightarrow b^2\ge\frac{4a^2+c^2}{2}=2a^2+\frac{c^2}{2}\)

\(\Rightarrow a^2+b^2+c^2\ge a^2+c^2+2a^2+\frac{c^2}{2}=3a^2+\frac{3}{2}c^2\) (1)

Mặt khác \(2< a+c\Rightarrow4< \left(a+c\right)^2=\left(\sqrt{\frac{1}{3}}.\sqrt{3}a+\sqrt{\frac{2}{3}}.\sqrt{\frac{3}{2}}c\right)^2\le\left(\frac{1}{3}+\frac{2}{3}\right)\left(3a^2+\frac{3}{2}c^2\right)\)

\(\Rightarrow3a^2+\frac{3}{2}c^2>4\) (2)

(1);(2) \(\Rightarrow a^2+b^2+c^2>4\) (đpcm)

Answer 2

[Việt Khải Võ]

thanks <3