\(a^2+\dfrac{1}{4}\ge a\)
\(b^2+\dfrac{1}{4}\ge b\)
\(c^2+\dfrac{1}{4}\ge c\)
Cộng theo vế suy ra đpcm
Xét hiệu:
\(a^2+b^2+c^2+\dfrac{3}{4}-a-b-c\)
\(=a^2+b^2+c^2+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}-a-b-c\)
\(=\left(a^2-a+\dfrac{1}{4}\right)+\left(b^2-b+\dfrac{1}{4}\right)+\left(c^2-c+\dfrac{1}{4}\right)\)
\(=\left(a-\dfrac{1}{2}\right)^2+\left(b-\dfrac{1}{2}\right)^2+\left(c-\dfrac{1}{2}\right)^2\ge0\) ( luôn đúng)
Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{2}\)
a2+b2+c2+34−a−b−ca2+b2+c2+34−a−b−c
=a2+b2+c2+14+14+14−a−b−c=a2+b2+c2+14+14+14−a−b−c
=(a2−a+14)+(b2−b+14)+(c2−c+14)=(a2−a+14)+(b2−b+14)+(c2−c+14)
=(a−12)2+(b−12)2+(c−12)2≥0=(a−12)2+(b−12)2+(c−12)2≥0 ( luôn đúng)
Dấu "=" xảy ra khi: a=b=c=12
