Question

Cho a,b,c là các số không âm. Chứng minh rằng:

a + b + c ≥ \(\sqrt{ab}\) + \(\sqrt{bc}\) + \(\sqrt{ca}\)

Answer 1

Ta có :

\(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

\(\Leftrightarrow2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

\(\Leftrightarrow2a+2b+2c-2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\ge0\)

\(\Leftrightarrow2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}\ge0\)

\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ca}+a\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\) ( luôn đúng )

Dấu '' = '' xảy ra khi a = b = c .

Answer 2