Ta cm \({1+4a \over 1-4a} \leq \dfrac{2a+b+c}{b+c} \Leftrightarrow (1+4a)(b+c) \leq (1-4a)(2a+b+c) \Leftrightarrow 4ab+4ac \leq a-4a^{2} \Leftrightarrow 4(a+b+c) \leq 1 (Luôn đúng)\)

Khi đó bđt cần cm tương đương \(\sum \dfrac{2a+b+c}{b+c} \leq 6+ \sum \dfrac{1}{a}. \ (\dfrac{\sqrt{b}-\sqrt{c}}{\sqrt{b}+\sqrt{c}}) ^2 \Leftrightarrow \sum (\dfrac{2a+b+c}{b+c} -2) \leq \sum \dfrac{1}{a}. \ (\dfrac{\sqrt{b}-\sqrt{c}}{\sqrt{b}+\sqrt{c}}) ^2 \Leftrightarrow \sum \dfrac{2a-b-c}{b+c} \leq \sum \dfrac{1}{a}. \ (\dfrac{\sqrt{b}-\sqrt{c}}{\sqrt{b}+\sqrt{c}}) ^2\)(*)

Thật vậy, có \(\sum \dfrac{2a-b-c}{b+c} = \sum \dfrac{a-b}{b+c} + \sum \dfrac{a-c}{b+c} = \sum (\dfrac{a-b}{b+c} + \dfrac{b-a}{c+a}) = \sum \dfrac{(a-b)^{2}}{(a+c)(b+c)}\)

\(\sum \dfrac{1}{a}. \ (\dfrac{\sqrt{b}-\sqrt{c}}{\sqrt{b}+\sqrt{c}}) ^2 = \sum \dfrac{1}{a}. \ \dfrac{(b-c)^2}{(\sqrt{b}+\sqrt{c})^4} = \sum \dfrac{(b-c)^2}{(\sqrt{ab}+\sqrt{ca})^2 .(\sqrt{b}+\sqrt{c})^2} \geq \sum \dfrac{(b-c)^2} {(c+a)(a+b)(\sqrt{b}+\sqrt{c})^2}\)

Khi đó bđt (*) tương đương \(\sum [ \dfrac{(b-c)^2}{(a+b)(c+a)}.(\dfrac{1}{(\sqrt{b}+\sqrt{c})^2} -1)] \geq 0\) (Luôn đúng do \(\dfrac{1}{2} >\dfrac{1}{4} \geq a+b+c>b+c \geq \dfrac{(\sqrt{b}+\sqrt{c})^2}{2} \Rightarrow \dfrac{1}{(\sqrt{b}+\sqrt{c})^2}-1>0\))

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