Lời giải:
Ta có:
\(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\leq \frac{3}{2}\)
\(\Leftrightarrow \frac{a-bc}{a(a+b+c)+bc}+\frac{b-ac}{b(a+b+c)+ca}+\frac{c-ab}{c(a+b+c)+ab}\leq \frac{3}{2}\)
\(\Leftrightarrow \frac{a-bc}{(a+b)(a+c)}+\frac{b-ac}{(b+a)(b+c)}+\frac{c-ab}{(c+a)(c+b)}\leq \frac{3}{2}\)
\(\Leftrightarrow \frac{(a-bc)(b+c)+(b-ac)(a+c)+(c-ab)(a+b)}{(a+b)(b+c)(c+a)}\leq \frac{3}{2}\)
\(\Leftrightarrow (a-bc)(b+c)+(b-ac)(a+c)+(c-ab)(a+b)\leq \frac{3}{2}(a+b)(b+c)(c+a)\)
\(\Leftrightarrow 2(ab+bc+ac)-[ab(a+b)+bc(b+c)+ac(a+c)]\leq \frac{3}{2}(1-a)(1-b)(1-c)\)
\(\Leftrightarrow 4(ab+bc+ac)-2[ab(a+b)+bc(b+c)+ac(c+a)]\leq 3(ab+bc+ac-abc)\)
\(\Leftrightarrow ab+bc+ac+3abc\leq 2[ab(a+b)+bc(b+c)+ca(c+a)]\)
\(\Leftrightarrow ab+bc+ac+9abc\leq 2[ab(a+b+c)+bc(a+b+c)+ac(a+b+c)]\)
\(\Leftrightarrow ab+bc+ac+9abc\leq 2(a+b+c)(ab+bc+ac)\)
\(\Leftrightarrow ab+bc+ac+9abc\leq 2(ab+bc+ac)\)
\(\Leftrightarrow 9abc\leq ab+bc+ac\)
\(\Leftrightarrow 9abc\leq (a+b+c)(ab+bc+ac)\)
BĐT trên luôn đúng do theo BĐT AM-GM ta có:
\((a+b+c)(ab+bc+ac)\geq 3\sqrt[3]{abc}.3\sqrt[3]{a^2b^2c^2}=9abc\)
Vậy ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)
