Question

[edition quan]

a²+b²+c²\(\ge\)\(\sqrt{abc}\).(\(\sqrt{a}+\sqrt{b}+\sqrt{c}\))

Answer 1

[Nguyễn Tất Đạt]

Hình như thiếu điều kiện a;b;c không âm

Áp dụng BĐT  Cauchy cho 2 số: \(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2ab+2bc+2ca\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Tiếp tục dùng Cauchy: \(ab+bc\ge2\sqrt{ab^2c}\)

Tương tự: \(bc+ca\ge2\sqrt{abc^2};ca+ab\ge2\sqrt{a^2bc}\)

\(\Rightarrow2\left(ab+bc+ca\right)\ge2\sqrt{a^2bc}+2\sqrt{ab^2c}+2\sqrt{abc^2}\)

\(\Leftrightarrow ab+bc+ca\ge\sqrt{a^2bc}+\sqrt{ab^2c}+\sqrt{abc^2}=\sqrt{abc}.\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrow a^2+b^2+c^2\ge\sqrt{abc}.\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Dấu "=" xảy ra <=> a=b=c.

Answer 2

[Đinh quang hiệp]

\(\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

thiếu a;b;c k âm nhá

áp dụng bđt cosi ngược ta có:

\(\sqrt{bc}< =\frac{b+c}{2}\Rightarrow a\sqrt{bc}< =\frac{a\left(b+c\right)}{2}=\frac{ab+ac}{2}\)

tương tự \(b\sqrt{ac}< =\frac{ab+bc}{2};c\sqrt{ab}< =\frac{ac+bc}{2}\)

\(\Rightarrow\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)=a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}< =\frac{ab+ac}{2}+\frac{ab+bc}{2}+\frac{ac+bc}{2}\)

\(=\frac{ab+ac+ab+bc+ac+bc}{2}=\frac{2ab+2bc+2ac}{2}=ab+ac+bc\left(1\right)\)

\(2ab< =a^2+b^2;2bc< =b^2+c^2;2ac< =a^2+c^2\Rightarrow2ab+2bc+2ac< =\)

\(a^2+b^2+b^2+c^2+a^2+c^2=2a^2+2b^2+2c^2\Rightarrow ab+bc+ac< =a^2+b^2+c^2\left(2\right)\)

từ \(\left(1\right)\left(2\right)\Rightarrow\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)< =a^2+b^2+c^2\)

\(\Rightarrow a^2+b^2+c^2>=\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\left(đpcm\right)\)

dấu = xảy ra khi a=b=c