Question

chứng minh a²+b²+c²>=2(a+b+c)-3 .Giúp mình vs nhé!

Answer 1

Ta có:

\(\left(a-1\right)^2\ge0\Leftrightarrow a^2-2a+1\ge0\Leftrightarrow a^2+1\ge2a\) (1)

\(\left(b-1\right)^2\ge0\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow b^2+1\ge2b\) (2)

\(\left(c-1\right)^2\ge0\Leftrightarrow c^2-2c+1\ge0\Leftrightarrow c^2+1\ge2c\) (3)

Từ (1), (2) và (3) suy ra;

\(a^2+1+b^2+1+c^2+1\ge2a+2b+2c\)

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

<=> \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\) \(\xrightarrow[]{}\) đpcm

Dấu "=" xảy ra khi a=b=c=1

Answer 2

Ta có: \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)

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

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

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

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