1 )
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a+1\ge2\sqrt{a}\\b+1\ge2\sqrt{b}\\a+c\ge2\sqrt{ac}\\b+c\ge2\sqrt{bc}\end{matrix}\right.\)
\(\Rightarrow VT\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{ac}.2\sqrt{bc}=16abc\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(a+c\right)\left(b+c\right)\ge16abc\)( đpcm )
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 a,b,c > 0 thõa mãn a+b+c=3
\(CMR:\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+a\right)\left(c+b\right)}\ge\dfrac{3}{4}\)
Chứng minh các BĐT sau:
a/ \(2\left(a^4+1\right)+\left(b^2+1\right)^2\ge2\left(ab+1\right)^2\)
b/ \(3\left(a^2+b^2\right)-ab+4\ge2\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\)
Cho a,b,c dương. Chứng minh
\(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(c+a\right)^2}\ge\dfrac{3\sqrt{3abc\left(a+b+c\right)}.\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)^3}\)
Giúp vs mọi người ơi
1. a,b,c > 0. C/m: \(\dfrac{c^2}{a+b}+\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}>=\dfrac{a+b+c}{2}\)
2. a,b,c > 0 và a+b+c <= 1. C/m: \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}>=9\)
3. a,b,c là 3 cạnh của một tam giác; \(p=\dfrac{a+b+c}{2}\)
C/m: \(\dfrac{1}{\left(p-a\right)^2}+\dfrac{1}{\left(p-b\right)^2}+\dfrac{1}{\left(p-c\right)^2}>=\dfrac{p}{\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
4. a,b,c > 0 và (a+c)(b+c)=1
C/m: \(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(a+c\right)^2}+\dfrac{1}{\left(b+c\right)^2}>=4\)
Cho 3 số dương a,b,c
CMR : \(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(a+c\right)^2}\ge\dfrac{9}{4\left(ab+ac+bc\right)}\)
với ∀a,b,c thuộc R, CMR:
\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\ge2+\frac{2\left(a+b+c\right)}{\sqrt[3]{abc}}\)
cho cac so thuc duong a b c thoa a^2+b^2+c^2>=3 chung minh
\(\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}+\frac{\left(b+1\right)\left(c+2\right)}{\left(c+1\right)\left(c+5\right)}+\frac{\left(c+1\right)\left(a+2\right)}{\left(a+1\right)\left(a+5\right)}\ge\frac{3}{2}\)
giúp mình với!! thanks nha^^
cho a, b, c > 0 thỏa mãn abc=1. cmr:\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{b^3}{c\left(a+1\right)}+\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3}{2}\)
