Question

Cho a , b , c > 0 . CMR : \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

Answer 1

\(a^2+b^2+c^2+3\) ≥ \(2\left(a+b+c\right)\)

⇔ \(a^2-2a+1+b^2-2b+1+c^2-2c+1\text{≥}0\)

⇔ \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\text{≥ }0\) ( luôn đúng )

\("="\text{⇔}a=b=c=1\)

Answer 2

Ta có :

\(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c\ge0\)

\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) ( Luôn đúng )

Dấu " = " xảy ra khi a = b = c = 1 .