Các câu hỏi tương tự
cho a;b;c >0 va a+b+c=6 c/m
\(\left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right)\left(1+\frac{1}{c^3}\right)>=\frac{729}{512}\)
21 tháng 8 2018 lúc 20:57
Cho a,b,c > 0 và a+b+c=6. Chứng minh rằng:
\(\left(1+\frac{1}{^{a^3}}\right)\left(1+\frac{1}{b^3}\right)\left(1+\frac{1}{c^3}\right)\ge\frac{729}{512}\)
Cho: a,b,c>0 thỏa mãn a+b+c=6
CMR \(\left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right)\left(1+\frac{1}{c^3}\right)\ge\frac{729}{512}\)
Cho a, b, c > 0 và a + b + c = 3. CMR: \(\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(b+1\right)\left(c+1\right)}+\frac{c^3}{\left(c+1\right)\left(a+1\right)}\ge\frac{3}{4}\)
Cho a;b;c là các số dương thỏa mãn a+b+c=\(\frac{3}{2}\). Chứng minh rằng:
B= \(\left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right)\left(1+\frac{1}{c^3}\right)\) ≥ 729
Cho a, b, c > 0 và abc = 1. CMR:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\)
cho a,b,c > 0 cmr: \(\frac{b^2a}{a^3\left(b+c\right)}+\frac{c^2a}{b^3\left(c+a\right)}+\frac{a^2b}{c^3\left(a+b\right)}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
cho a,b,c>0 va a+b+c=1. CMR
\(\frac{a}{\left(b+c\right)^3}+\frac{b}{\left(a+c\right)^3}+\frac{c}{\left(a+b\right)^3}\ge\frac{27}{8\left(a+b+c\right)^2}\)
\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b^3}{\left(a+1\right)\left(c+1\right)}+\frac{c^3}{\left(a+1\right)\left(b+1\right)}\\ \\ \)Cho a,b,c > 0 và a+b+c=3 Chứng minh rằng :