Question

cho 3 số dương thỏa mãn \(a+b+c\le\dfrac{3}{2}\). tìm GTNN của :

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

Answer 1

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

\(=\dfrac{1}{\sqrt{1^2+4^2}}.\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1^2+4^2\right)}+\dfrac{1}{\sqrt{1^2+4^2}}.\sqrt{\left(b^2+\dfrac{1}{c^2}\right)\left(1^2+4^2\right)}+\dfrac{1}{\sqrt{1^2+4^2}}.\sqrt{\left(c^2+\dfrac{1}{a^2}\right)\left(1^2+4^2\right)}\)\(=\dfrac{1}{\sqrt{17}}.\left[\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1^2+4^2\right)}+\sqrt{\left(b^2+\dfrac{1}{c^2}\right)\left(1^2+4^2\right)}+\sqrt{\left(c^2+\dfrac{1}{a^2}\right)\left(1^2+4^2\right)}\right]\)

\(\ge\dfrac{1}{\sqrt{17}}.\left[\left(a.1+\dfrac{1}{b}.4\right)+\left(b.1+\dfrac{1}{c}.4\right)+\left(c.1+\dfrac{1}{a}.4\right)\right]\)

\(=\dfrac{1}{\sqrt{17}}.\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\)

\(=\dfrac{1}{\sqrt{17}}.\left[\left(a+\dfrac{1}{4a}\right)+\left(b+\dfrac{1}{4b}\right)+\left(c+\dfrac{1}{4c}\right)+\dfrac{15}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)

\(\ge\dfrac{1}{\sqrt{17}}.\left(1+1+1+\dfrac{15}{4}.\dfrac{9}{a+b+c}\right)\)

\(\ge\dfrac{1}{\sqrt{17}}.\left(3+\dfrac{15}{4}.\dfrac{9}{\dfrac{3}{2}}\right)\)

\(=\dfrac{1}{\sqrt{17}}.\dfrac{51}{2}=\dfrac{3\sqrt{17}}{2}\)

- Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)

- Vậy \(MinA=\dfrac{3\sqrt{17}}{2}\)

Answer 2

Ta có \(A\sqrt{\dfrac{17}{4}}=\sqrt{\dfrac{17}{4}\left(a^2+\dfrac{1}{b^2}\right)}+\sqrt{\dfrac{17}{4}\left(b^2+\dfrac{1}{c^2}\right)}+\sqrt{\dfrac{17}{4}\left(c^2+\dfrac{1}{a^2}\right)}\)

Có :\(\sqrt{\dfrac{17}{4}\left(a^2+\dfrac{1}{b^2}\right)}=\sqrt{\left[2^2+\left(\dfrac{1}{2}\right)^2\right]\left[\left(\dfrac{1}{b}\right)^2+a^2\right]}\ge\sqrt{\left(\dfrac{2}{b}+\dfrac{a}{2}\right)^2}\)

= \(\dfrac{2}{b}+\dfrac{a}{2}\)

Tương tự đươc \(\dfrac{A\sqrt{17}}{2}\ge\dfrac{2}{b}+\dfrac{a}{2}+\dfrac{2}{c}+\dfrac{b}{2}+\dfrac{2}{a}+\dfrac{c}{2}\)

\(=\left(\dfrac{a}{2}+\dfrac{1}{8a}\right)+\left(\dfrac{b}{2}+\dfrac{1}{8b}\right)+\left(\dfrac{c}{2}+\dfrac{1}{8c}\right)+\dfrac{15}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\ge2\sqrt{\dfrac{a}{2}.\dfrac{1}{8a}}+2\sqrt{\dfrac{b}{2}.\dfrac{1}{8b}}+2\sqrt{\dfrac{c}{2}.\dfrac{1}{8c}}+\dfrac{15}{8}.\dfrac{9}{a+b+c}\)

\(\ge\dfrac{3}{2}+\dfrac{15}{8}.\dfrac{9}{\dfrac{3}{2}}=\dfrac{51}{4}\Leftrightarrow A\ge\dfrac{3\sqrt{17}}{2}\)

"=" KHI a = b = c = 1/2