Question

Cho 3 số thực a,b,c .Chứng minh rằng :

\(\dfrac{2a^3}{a^6+bc}+\dfrac{2b^3}{b^6+ac}+\dfrac{2c^3}{c^6+ab}\le\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\)

Answer 1

Ta có: \(a^2+b^2+c^2\ge ab+bc+ca\ge\sqrt[]{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Do đó:

\(VT\le\dfrac{2a^3}{2\sqrt{a^6bc}}+\dfrac{2b^3}{2\sqrt{b^6ac}}+\dfrac{2c^3}{2\sqrt{c^3ab}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=\dfrac{\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}{abc}\)

\(\le\dfrac{a^2+b^2+c^2}{abc}=\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\)

Dấu "=" xảy ra khi \(a=b=c=1\)