Question

Cho a, b, c không âm. Chứng minh

√(a+b) + √(b+c) + √(c+a) >= √(2a) + √(2b) + √(2c)

Answer 1

Với 2 số dương bất kì ta có:

\(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\Leftrightarrow2x+2y\ge\left(\sqrt{x}+\sqrt{y}\right)^2\Leftrightarrow x+y\ge\frac{1}{2}\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(\Leftrightarrow\sqrt{x+y}\ge\frac{\sqrt{2}}{2}\left(\sqrt{x}+\sqrt{y}\right)\)

Áp dụng:

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

Dấu "=" xảy ra khi \(a=b=c\)