Lời giải:
Ta có:
\(A=\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}\)
\(\Rightarrow A+\frac{29}{2}=\frac{4a}{b+c-a}+2+\frac{9b}{a+c-b}+\frac{9}{2}+\frac{16c}{a+b-c}+8\)
\(A+\frac{29}{2}=\frac{2(a+b+c)}{b+c-a}+\frac{\frac{9}{2}(a+b+c)}{a+c-b}+\frac{8(a+b+c)}{a+b-c}\)
\(A+\frac{29}{2}=(a+b+c)\left(\frac{2}{b+c-a}+\frac{\frac{9}{2}}{a+c-b}+\frac{8}{a+b-c}\right)\)
\(\geq (a+b+c).\frac{(\sqrt{2}+\sqrt{\frac{9}{2}}+\sqrt{8})^2}{b+c-a+a+c-b+a+b-c}=\frac{81}{2}\)
(Áp dụng BĐT S.Vac -xơ)
\(\Rightarrow A\geq 26\)
Vậy \(A_{\min}=26\)
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