ta có \(\left(a+b\right)^2\ge4ab\Rightarrow1\ge4ab\Leftrightarrow ab\le\frac{1}{4}\)
\(\Rightarrow\frac{1}{4ab}\ge1\Rightarrow\frac{8}{4ab}\ge8\) hay \(\frac{2}{ab}\ge8\)
ta có
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\\ =1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}\\ =1+\frac{a+b+1}{ab}\\ =1+\frac{2}{ab}\ge1+8=9\)
(đpcm)
Cho a, b, c là các số dương. Chứng minh: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
CMR \(\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)>=\left(\frac{10}{3}\right)^2\) voi a,b,c >0 va a+b+c=1
Cho \(a,b,c\ge1\). CMR:
\(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge9\)
Cho số thực dương a,b,c thỏa mãn \(a+b+c\le3\). CMR
\(a+b+c+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
cho a,b > 0 thỏa mãn a + b = 2. Cmr: \(2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge10\)
Cho a,b,c dương và abc=1
CMR: \(\frac{a^4}{2\left(b+c\right)^2}+\frac{b^4}{2\left(a+c\right)^2}+\frac{c^4}{2\left(a+b\right)^2}+\frac{1}{c^2\left(a+c\right)\left(a+b\right)}+\frac{1}{b^2\left(a+b\right)\left(b+c\right)}+\frac{1}{a^2\left(a+c\right)\left(a+b\right)}\ge\frac{1}{8}\)
Cho a>0, b>0. CMR:
\(\frac{\left(a+1\right)^2}{b}+\frac{\left(b+1\right)^2}{a}\ge8\)
cho 0<a,b,c<\(\frac{1}{2}\)thỏa mãn a+b+c=1
CMR: \(\frac{1}{a\left(2b+2c-1\right)}+\frac{1}{b\left(2c+2a-1\right)}+\frac{1}{c\left(2a+2b-1\right)}\ge27\)
