Chủ đề này có 2 trả lời

[raeunho]

Cho a,b,c>0 thỏa $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. CMR $\frac{a^{2}}{a+bc}+\frac{b^{2}}{b+ca}+\frac{c^{2}}{c+ab}\geq \frac{a+b+c}{4}$

[Diepnguyencva]

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ac=abc; \frac{a^{2}}{a+bc}=\frac{a^{3}}{(a+b)(a+c)};$

Áp dụng Cauchy 3 số, $\frac{a^{3}}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq \frac{a}{2}$

[PugMath]

$VT=\sum{\frac{a^3}{a^2+ab+bc+ac}}=\sum{\frac{a^3}{(a+b)(a+c)}}+\sum{\frac{a+b}{8}}+\sum{\frac{a+c}{8}}-\sum{\frac{a+b}{4}}\geqslant\sum{\frac{3a}{4}}-\frac{2a+2b+2c}{4}=VP$