Question

[Trần Anh Thơ]

Cho các số thực dương a, b, c thỏa mãn cba + + ≥ 6 . Tìm GTNN của biểu thức:

A = \(\sqrt{a^2+\frac{1}{b+c}}+\sqrt{b^2+\frac{1}{a+c}}\sqrt{c^2+\frac{1}{a+b}}\)

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

Xét \(\left(a^2+\frac{1}{b+c}\right)\left(4^2+1^2\right)\ge\left(4a+\frac{1}{\sqrt{b+c}}\right)^2\)

=> \(\sqrt{a^2+\frac{1}{b+c}}\ge\frac{4a+\frac{1}{\sqrt{b+c}}}{\sqrt{17}}\)

Tương tự => \(\left\{{}\begin{matrix}\sqrt{b^2+\frac{1}{c+a}}\ge\frac{4b+\frac{1}{\sqrt{c+a}}}{\sqrt{17}}\\\sqrt{c^2+\frac{1}{a+b}}\ge\frac{4c+\frac{1}{\sqrt{a+b}}}{\sqrt{17}}\end{matrix}\right.\)

=> A \(\ge\frac{4\left(a+b+c\right)+\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}}{\sqrt{17}}\)

Có \(\frac{1}{\sqrt{a+b}}=\frac{4}{4.\sqrt{a+b}}\)

Mà \(\sqrt{\left(a+b\right).4}\le\frac{a+b+4}{2}\) => \(4\sqrt{a+b}\le a+b+4\)

=> \(\frac{1}{\sqrt{a+b}}\ge\frac{4}{a+b+4}\)

Tương tự => \(\left\{{}\begin{matrix}\frac{1}{\sqrt{b+c}}\ge\frac{4}{b+c+4}\\\frac{1}{\sqrt{c+a}}\ge\frac{4}{c+a+4}\end{matrix}\right.\)

=> \(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\) \(\ge4.\left(\frac{1}{b+c+4}+\frac{1}{c+a+4}+\frac{1}{a+b+4}\right)\)

\(\ge4.\frac{9}{2a+2b+2c+12}=\frac{3}{2}\)

=> \(A\ge\frac{4.6+\frac{3}{2}}{\sqrt{17}}=\frac{3.\sqrt{17}}{2}\)