\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)\ge a.2bc+b.2ca+c.2ab=2abc+2abc+2abc=6abc\)

Lời giải:
Ta có:

Nhân cả hai vế với $a+b+c$ , BĐT cần chứng minh tương đương với:

\(\frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)\)

\(\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)\)

\(\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2\)

\(\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2\)

\(\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\)

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Áp dụng BĐT Cauchy- Schwarz:

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \frac{9}{2(a+b+c)}\)

\(\Rightarrow a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\geq a^2+b^2+c^2+\frac{9abc}{a+b+c}\)

Ta cần chứng minh \(a^2+b^2+c^2+\frac{9abc}{a+b+c}\geq 2(ab+bc+ac)\)

\(\Leftrightarrow (a^2+b^2+c^2)(a+b+c)+9abc\geq 2(ab+bc+ac)(a+b+c)\)

\(\Leftrightarrow a^3+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(a+c)\)

(luôn đúng theo BĐT Schur)

Do đó ta có đpcm.

Dấu bằng xảy ra khi $a=b=c$

Chia 2 vế của BĐT cho \(\left(a+b+c\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\)L

\(3(a^2+b^2+c^2)(a+b)(b+c)(c+a)\ge(a+b+c)\left(\sum_{cyc}(a^2+b^2)(c+a)(c+b)\right)\)

\(\Leftrightarrow\sum_{perms}a^2b(a-b)^2\ge0\) *đúng* XD