Question

[Phùng Gia Bảo]

Cho ba số thực dương a,b,c thỏa \(ab+bc+ca=abc\). Chứng minh rằng: \(\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\)

Answer 1

[Darlingg🥝]

Áp dụng BĐT Bunhiacopxky ta có:

\(\left(a^2+2c^2\right)\left(1+2\right)\ge\left(a+2c^2\right)\)

\(\Rightarrow\sqrt{a^2+2c^2}\ge\frac{a+2c}{3}\)

\(\Rightarrow\frac{\sqrt{a^2+2c^2}}{ac}\ge\frac{a+2c}{\sqrt{3ac}}=\frac{ab+2bc}{\sqrt{3abc}}\)

\(\Rightarrow\hept{\begin{cases}\frac{\sqrt{c^2+2b^2}}{bc}\ge\frac{ac+2ab}{\sqrt{3abc}}\\\frac{\sqrt{b^2+2a^2}}{ab}\ge\frac{bc+2ac}{\sqrt{abc}}\end{cases}}\)

Ta được BĐT:

\(VT\ge\frac{1}{3}.\frac{ab+2abc+ac+2ab+bc+2ac}{abc}=\frac{1}{3}.\frac{3\left(ab+bc+ac\right)}{abc}\)

\(=\frac{1}{\sqrt{3}}.\frac{3abc}{abc}=3\)

=> đpcm

P/S: Làm tắt vs đoạn này k^o chắc mấy :V

Answer 2

[Nyatmax]

Repair đề \(\Sigma_{cyc}\frac{\sqrt{2a^2+b^2}}{ab}\ge3\sqrt{3}\).Because dấu '=' xảy ra khi \(a=b=c=3\)

Không use condition của đề bài :))

Ta co:

\(VT=\sqrt{\frac{a}{b}+\frac{a}{b}+\frac{b}{a}}+\sqrt{\frac{b}{c}+\frac{b}{c}+\frac{c}{b}}+\sqrt{\frac{c}{a}+\frac{c}{a}+\frac{a}{c}}\)

\(\Rightarrow VT\ge\sqrt{3\sqrt[3]{\frac{a}{b}}}+\sqrt{3\sqrt[3]{\frac{b}{c}}}+\sqrt{3\sqrt[3]{\frac{c}{a}}}\ge3\sqrt[3]{\sqrt{3\sqrt[3]{\frac{a}{b}}.\sqrt{3\sqrt[3]{\frac{b}{c}}.\sqrt{3\sqrt[3]{\frac{c}{a}}}}}}=3\sqrt{3}\)

equelity iff \(a=b=c=3\)