\(a^2+b^2+c^2+3\)≥2(a+b+c)
<=>\(a^2+b^2+c^2+3-2a-2b-2c\)≥0
<=>\(a^2-2a+1+b^2-2b+1+c^2-2c+1\)≥0
<=>\(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\)≥0
=>\(a^2+b^2+c^2+3\)≥2(a+b+c+
<=>a 2+b 2+c2+3-2a-2b-2c≥0
<=>a 2-2a+1+b2-2b+1+c 2-2c+1≥0
<=>(a-1) 2+(b-1) 2+(c-1) 2≥0 Luôn đúng
<=>a 2+b 2+c2≥2(a+b+c)(đpcm)