Question

cho a,b,c>0 và a+b+c=3 cmr

\(a^3+b^3+c^3+\dfrac{15}{4}abc\ge\dfrac{27}{4}\)

Answer 1

Lời giải:
Ta có:

\(a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)\)

\(=27-3(3-a)(3-b)(3-c)\)

\(=27-3[27-9(a+b+c)+3(ab+bc+ac)-abc]\)

\(=27-3[3(ab+bc+ac)-abc]=27-9(ab+bc+ac)+3abc\)

Do đó:

\(A=a^3+b^3+c^3+\frac{15}{4}abc=27-9(ab+bc+ac)+\frac{27}{4}abc(*)\)

Áp dụng BĐT Schur :

\(abc\geq (a+b-c)(b+c-a)(c+a-b)\)

\(\Leftrightarrow abc\geq (3-2a)(3-2b)(3-2c)\)

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

\(\Leftrightarrow 9abc\geq 12(ab+bc+ac)-27\)

\(\Leftrightarrow 3abc\geq 4(ab+bc+ac)-9\)

\(\Rightarrow \frac{27}{4}abc\geq 9(ab+bc+ac)-\frac{81}{4}(**)\)

Từ \((*); (**)\Rightarrow A\geq 27-\frac{81}{4}=\frac{27}{4}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Answer 2

Ta có:

a³+b³+