\(VT\ge\sum\left(\dfrac{a^3}{2a+b+c}\right)=\sum\left(\dfrac{a^3}{\sum a+a}\right)=\sum\dfrac{a^3}{3+a}\)
Ta có BĐT phụ :
\(\dfrac{a^3}{a+3}\ge\dfrac{11a-7}{16}\)(*)
\(\Leftrightarrow\left(16a+21\right)\left(a-1\right)^2\ge0\) (luôn đúng với mọi a>0)
Áp dụng BĐT (*) ta có :
\(\sum\dfrac{a^3}{3+a}\ge\dfrac{11\sum a-21}{16}=\dfrac{33-21}{16}=\dfrac{12}{16}=\dfrac{3}{4}\)
cho a,b,c >0 thõa mãn abc = 1
\(CMR:\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(a+b\right)}\ge\dfrac{3}{4}\)
Cho 3 số dương a,b,c
CMR : \(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(a+c\right)^2}\ge\dfrac{9}{4\left(ab+ac+bc\right)}\)
Cho a,b,c là số dương. CMR:
1. \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
2. \(a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\le a^3+b^3+c^3\)
3. \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
CMR : a,b,c >0
\(\left(a^3+b^3+c^3\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\dfrac{>}{ }\left(a+b+c\right)^2\)
Cho a,b,c dương. Chứng minh
\(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(c+a\right)^2}\ge\dfrac{3\sqrt{3abc\left(a+b+c\right)}.\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)^3}\)
Cho a,b,c,d là số dương. Cmr
a/ \(\left(\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{d^3}+\dfrac{d}{a^3}\right)\left(a+b\right)\left(b+c\right)\ge16\)
b/ \(\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
giúp mình với!! thanks nha^^
cho a, b, c > 0 thỏa mãn abc=1. cmr:\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{b^3}{c\left(a+1\right)}+\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3}{2}\)
Cho a , b , c là các số thực dương . Chứng minh rằng
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Cho a,b,c >0 thõa mãn a+b+c = 3
\(CMR:\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)
