Question

Với a,b,c là 3 cạnh tam giác hãy tìm GTNN của biểu thức: P = \(\dfrac{4a}{b+c-a}+\dfrac{9b}{a+c-b}+\dfrac{16c}{a+b-c}\)

Answer 1

Lời giải:

Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}\)

\(P+\frac{29}{2}=\frac{4a}{b+c-a}+2+\frac{9b}{a+c-b}+\frac{9}{2}+\frac{16c}{a+b-c}+8\)

\(=\frac{2(a+b+c)}{b+c-a}+\frac{9(a+b+c)}{2(a+c-b)}+\frac{8(a+b+c)}{a+b-c}\)

\(=2(a+b+c)\left(\frac{1}{b+c-a}+\frac{\frac{9}{4}}{a+c-b}+\frac{4}{a+b-c}\right)\)

\(\geq 2(a+b+c).\frac{(1+\frac{3}{2}+2)^2}{b+c-a+a+c-b+a+b-c}=\frac{81}{2}.(a+b+c).\frac{1}{a+b+c}=\frac{81}{2}\)

\(\Rightarrow P\geq \frac{81}{2}-\frac{29}{2}=26\)

Vậy \(P_{\min}=26\)