Question

[nguyễn hương mây]

C/m căn(a^2 + b^2) + căn(b^2 + c^2) + căn(c^2 + a^2`) >= căn 2 * (a+b+c) với mọi a, b, c

Answer 1

[Nguyễn Việt Lâm]

Với mọi a;b ta luôn có:

\(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\)

\(\Leftrightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\)

\(\Rightarrow\sqrt{a^2+b^2}\ge\sqrt{\dfrac{1}{2}\left(a+b\right)^2}=\dfrac{\sqrt{2}}{2}\left|a+b\right|\ge\dfrac{\sqrt{2}}{2}\left(a+b\right)\)

Tương tự:

\(\sqrt{b^2+c^2}\ge\dfrac{\sqrt{2}}{2}\left(b+c\right)\) ; \(\sqrt{c^2+a^2}\ge\dfrac{\sqrt{2}}{2}\left(c+a\right)\)

Cộng vế:

\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\sqrt{2}\left(a+b+c\right)\)

Dấu "=" xảy ra khi \(a=b=c\ge0\)