Chủ đề này có 2 trả lời

[eatchuoi19999]

Cho $a,b,c>0$ thỏa mãn $ab+bc+ca=3.$ Chứng minh: $\frac{1}{b(b+a)}+\frac{1}{c(c+b)}+\frac{1}{a(a+c)}\geqslant \frac{3}{2}.$

 

[Hoang Tung 126]

Theo bđt AM-GM có :$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(a+c)}\geq 3\sqrt[3]{\frac{1}{abc(a+b)(b+c)(c+a)}}=\frac{3}{\sqrt[3]{(ab+ac)(bc+ba)(ca+cb)}}\geq \frac{3}{\frac{2(ab+bc+ac)}{3}}=\frac{9}{2(ab+bc+ac)}=\frac{9}{6}=\frac{3}{2}$(đpcm)

[Trang Luong]

Cho $a,b,c>0$ thỏa mãn $ab+bc+ca=3.$ Chứng minh: $\frac{1}{b(b+a)}+\frac{1}{c(c+b)}+\frac{1}{a(a+c)}\geqslant \frac{3}{2}.$

Ta có : $\frac{1}{b(b+a)}+\frac{1}{c(c+b)}+\frac{1}{a(a+c)}= \frac{c}{b(ac+bc)}+\frac{a}{c(ac+ba)}+\frac{b}{a(ba+bc)}=\frac{c}{b(3-ab)}+\frac{a}{c(3-bc)}+\frac{b}{a(3-ac)}\geq 3.\sqrt[3]{\frac{1}{\left ( 3-ab \right )\left ( 3-ac \right )\left ( 3-bc \right )}}\geq \frac{3}{\left ( \frac{3-ab+3-bc+3-ac}{3} \right )}=\frac{3}{2}$