Question

Cho (a+b)(b+c)(c+a)=1. Các số a, b, c > 0.

CM: \(ab+bc+ca\le\dfrac{3}{4}\)

Answer 1

Từ \(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{8\left(a+b+c\right)^3}{27}\Rightarrow a+b+c\ge\dfrac{3}{2}\)

Áp dụng bổ đề \((a+b)(b+c)(c+a)\geq \frac{8}{9}(a+b+c)(ab+bc+ca)\)

\(1\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\ge\dfrac{8}{9}\cdot\dfrac{3}{2}\left(ab+bc+ca\right)\)

\(=\dfrac{4}{3}\left(ab+bc+ca\right)\Rightarrow ab+bc+ca\le\dfrac{3}{4}\)

Answer 2

Bổ đề(tự cm): 8(a+b+c)(ab+bc+ca) \(\le\)9(a+b)(b+c)(c+a)

Từ đó suy ra \(ab+bc+ca\le\dfrac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\left(a+b+c\right)}=\dfrac{9}{4\left(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right)}=\dfrac{9}{4.3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\dfrac{9}{4.3}=\dfrac{3}{4}\)