Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH
Các câu hỏi tương tự
Cho a,b,c > 0 thõa mãn a+b+c=3
\(CMR:\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+a\right)\left(c+b\right)}\ge\dfrac{3}{4}\)
CMR : a,b,c >0
\(\left(a^3+b^3+c^3\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\dfrac{>}{ }\left(a+b+c\right)^2\)
2 tháng 9 2019 lúc 13:16
Cho a,b,c>0.CMR:
\(\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}+\frac{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+c\right)^3}\ge16\)
Help!
cho a,b,c >0 thõa mãn abc = 1
\(CMR:\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(a+b\right)}\ge\dfrac{3}{4}\)
Bài 3. Cho \(a,b,c\in R\). Chứng minh các bất đẳng thức sau:
\(a,\frac{a^2+3}{\sqrt{a^2+2}}>2\)
\(b,\left(a^5+b^5\right)\left(a+b\right)\ge\left(a^4+b^4\right)\left(a^2+b^2\right)\) \(\left(ab>0\right)\)
\(c,\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
cho a,b,c > 0 thỏa mãn \(a^2+b^2+c^2=3\) . Cmr:
\(\left(\frac{4}{a^2+b^2}+1\right)\left(\frac{4}{b^2+c^2}+1\right)\left(\frac{4}{c^2+a^2}+1\right)\ge3\left(a^2+b^2+c^2\right)\)
10 tháng 2 2021 lúc 20:14
Cho 3 số dương a,b,c
CMR : \(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(a+c\right)^2}\ge\dfrac{9}{4\left(ab+ac+bc\right)}\)
cmr \(2\left(a^2+b^2\right)>=\left(a+b\right)^2\)
\(3\left(a^2+b^2+c^2\right)>=\left(a+b+c\right)^2\)
\(\left(a+b+c\right)^2>=3\left(ab+bc+ca\right)\)
a)cho a,b,c > 0 . Cmr: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
b)cho a,b,c > 0 thỏa mãn \(a^2+b^2+c^2=3\) . Cmr: \(\left(a^2b+b^2c+c^2a\right)\left(a+b+c\right)\ge9abc\)