Question

Cho a+b+c=3 (a,b,c > 0) CMR: a²+b²+c²+abc\(\ge\)4

Answer 1

Lời giải:

Theo BĐT Schur bậc 3:

\(abc\geq (a+b-c)(b+c-a)(c+a-b)=(3-2a)(3-2b)(3-2c)\)

\(\Leftrightarrow abc\geq 27+12(ab+bc+ac)-18(a+b+c)-8abc=-27+12(ab+bc+ac)-8abc\)

\(\Rightarrow 9abc\geq 12(ab+bc+ac)-27\Rightarrow abc\geq \frac{4}{3}(ab+bc+ac)-3\)

Do đó:

\(a^2+b^2+c^2+abc\geq a^2+b^2+c^2+\frac{4}{3}(ab+bc+ac)-3\)

\(=(a+b+c)^2-\frac{2}{3}(ab+bc+ac)-3=6-\frac{2}{3}(ab+bc+ac)\)

Mặt khác theo hệ quả quen thuộc của BĐT AM-GM:
\(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\)

\(\Rightarrow a^2+b^2+c^2+abc\geq 6-\frac{2}{3}(ab+bc+ac)\geq 6-\frac{2}{3}.3=4\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$