Question

Cho a,b,c là các số thực thỏa mãn \(\dfrac{1}{2}\)≤ a,b,c ≤\(1\)

Tìm GTNN của \(\dfrac{a}{b+c+1}+\dfrac{b}{a+c+1}+\dfrac{c}{b+a+1}\)

Answer 1

Lời giải:

Đặt \(P=\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}\)

\(P+3=\frac{a+b+c+1}{b+c+1}+\frac{b+a+c+1}{a+c+1}+\frac{c+b+a+1}{b+a+1}\)

\(=(a+b+c+1)\left(\frac{1}{b+c+1}+\frac{1}{a+c+1}+\frac{1}{b+a+1}\right)\)

Áp dụng BĐT Cauchy-Schwarz:

\(P+3\geq (a+b+c+1).\frac{9}{b+c+1+a+c+1+b+a+1}=\frac{9(a+b+c+1)}{2(a+b+c+1)+1}\)

Đặt \(a+b+c+1=t\). Vì \(a,b,c\geq \frac{1}{2}\Rightarrow t\geq \frac{5}{2}\)

Khi đó:

\(\frac{9(a+b+c+1)}{2(a+b+c+1)+1}=\frac{9t}{2t+1}=\frac{9}{2}-\frac{9}{2(2t+1)}\geq \frac{9}{2}-\frac{9}{2.(2.\frac{5}{2}+1)}=\frac{15}{4}\)

\(\Rightarrow P+3\geq \frac{9(a+b+c+1)}{2(a+b+c+1)+1}\geq \frac{15}{4}\)

\(\Rightarrow P\ge \frac{3}{4}\)

Vậy \(P_{\min}=\frac{3}{4}\Leftrightarrow a=b=c=\frac{1}{2}\)