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[dinhvu]

Cho $a,b,c$ thực dương thỏa mãn$(a+b)(b+c)(c+a)=8$. CMR 

$(\frac{a+b}{a+c}+\frac{a+c}{a+b}-1)(\frac{b+c}{b+a}+\frac{b+a}{b+c}-1)(\frac{c+b}{c+a}+\frac{c+a}{c+b}-1)\geq \frac{8abc}{(a^2+bc)(b^2+ac)(c^2+ab)}$

[truongphat266]

Best Answer

Cho $a,b,c$ thực dương thỏa mãn$(a+b)(b+c)(c+a)=8$. CMR 

$(\frac{a+b}{a+c}+\frac{a+c}{a+b}-1)(\frac{b+c}{b+a}+\frac{b+a}{b+c}-1)(\frac{c+b}{c+a}+\frac{c+a}{c+b}-1)\geq \frac{8abc}{(a^2+bc)(b^2+ac)(c^2+ab)}$

Ta có bổ đề sau: Cho $a,b,c \geq 0$ thì có: $$\frac{1}{(c+a)^2}+\frac{1}{(b+a)^2}\geq \frac{1}{a^2+bc}\Leftrightarrow \frac{(a^2-bc)^2+bc(b-c)^2}{(a+b)^2(a+c)^2(a^2+bc)}\geq 0.$$

Áp dụng bổ đề:

$$VT=\prod _{cyc}(a+b)(a+c)(\frac{1}{(c+a)^2}+\frac{1}{(b+a)^2}-\frac{1}{(c+a)(b+c)})\geq \prod_{cyc} \frac{a(b+c)}{a^2+bc}=VP.$$

Có điều phải chứng minh!

Edited by truongphat266, 17-09-2023 - 09:16.