Question

bài 1:cho ab+bc+3ca=1.Tìm min S=\(a^2+b^2+c^2\)

Answer 1

\(ab=\dfrac{1}{k}.a.kb\le\dfrac{1}{2k}\left(a^2+k^2b^2\right)\) , \(bc=\dfrac{1}{k}.c.kb\le\dfrac{1}{2k}\left(c^2+k^2b^2\right)\), \(3ac\le\dfrac{3}{2}\left(a^2+c^2\right)\)

\(\Rightarrow ab+cb+3ac\le a^2\left(\dfrac{1}{2k}+\dfrac{3}{2}\right)+c^2\left(\dfrac{1}{2k}+\dfrac{3}{2}\right)+b^2.k\)

Tìm k sao cho \(k=\dfrac{1}{2k}+\dfrac{3}{2}\). Khi đó \(a^2+b^2+c^2\ge\dfrac{1}{k}\)

Tìm ra \(k=\dfrac{3+\sqrt{17}}{4}\).Vậy \(S_{Min}=\dfrac{4}{3+\sqrt{17}}\)