Cho a,b,c dương thỏa mãn a+b+c=6.Tìm GTNN của P= $\frac{a}{4+b^2}$ +$\frac{c}{4+c^2}$+ $\frac{b}{4+a^2}$

Đề: Cho \(a,b,c\) dương thỏa mãn \(a+b+c=6\). Tìm GTNN của: \(P=\dfrac{a}{4+b^2}+\dfrac{b}{4+c^2}+\dfrac{c}{4+a^2}\)

Xét: \(\dfrac{4a}{4+b^2}=\dfrac{4a+ab^2 - ab^2}{4+b^2}=\dfrac{a(4+b^2)-ab^2}{4+b^2}=a-\dfrac{ab^2}{4+b^2}\)

Theo Cauchy có: \(4+b^2\ge 4b\)

\(\to \dfrac{4a}{4+b^2}\ge a-\dfrac{ab^2}{4b}=a-\dfrac{ab}{4}\)

Tương tự: \(\dfrac{4b}{4+c^2}\ge b-\dfrac{bc}{4},\dfrac{4c}{4+a^2}\ge c-\dfrac{ac}{4}\)

\(\to 4P\ge a+b+c-\dfrac{ab+bc+ac}{4}\)

Theo Cauchy có: \(a^2+b^2\ge 2ab,b^2+c^2\ge 2bc,c^2+a^2\ge 2ac\)

\(\to a^2+b^2+c^2\ge ab+bc+ac\\\to (a+b+c)^2\ge 3(ab+bc+ac)\\\to ab+bc+ac\le \dfrac{(a+b+c)^2}{3}\\\to 4P\ge a+b+c-\dfrac{(a+b+c)^2}{12}\\\to 4P\ge 6-\dfrac{6^2}{12}\\\to P\ge \dfrac{3}{4}\)

Dấu "=" xảy ra khi: \(a=b=c=2\)