Violympic toán 9
Các câu hỏi tương tự
CMR: \(\frac{2a^3+1}{4b\left(a-b\right)}\ge3\) \(\forall\left\{{}\begin{matrix}a\ge\frac{1}{2}\\\frac{a}{b}>1\end{matrix}\right.\)
Cho a,b,c>0 , chứng minh rằng:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
Cho \(\left\{{}\begin{matrix}a,b,c>0\\\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=3\end{matrix}\right.\)
Tìm MAX A \(=\frac{1}{\left(2a+b+c\right)^2}+\frac{1}{\left(2b+a+c\right)^2}+\frac{1}{\left(2c+a+b\right)^2}\)
\(\left(\frac{a^2-ab}{a^2b+b^3}-\frac{2a^2}{b^3-ab^2+a^2b-a3}\right)\cdot\left(1-\frac{b-1}{a}-\frac{b}{a^2}\right)=\frac{a+1}{ab}\)
17 tháng 11 2019 lúc 17:52
1. \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) Cmr: \(\frac{x^2}{\left(x+1\right)^2}+\frac{y^2}{\left(y+1\right)^2}+\frac{z^2}{\left(z+1\right)^2}\ge\frac{3}{4}\)\
2. \(a,b,c>0.\) cmr: \(\Sigma\frac{a^3}{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\le\frac{1}{a+b+c}\)
20 tháng 6 2020 lúc 22:34
Cho 3 số thực dương \(a;b;c\) thỏa mãn: \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2019\)
Tìm giá trị lớn nhất của biểu thức \(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
cho \(a,b,c>\frac{1}{2}\) và thỏa mãn \(a+b+c=3\).Chứng minh rằng
\(\frac{a^2}{\sqrt{5-2\left(b+c\right)}}+\frac{b^2}{\sqrt{5-2\left(a+c\right)}}+\frac{c^2}{\sqrt{5-2\left(a+b\right)}}\ge3\)
cho 0<a,b,c<\(\frac{1}{2}\)thỏa mãn a+b+c=1
CMR: \(\frac{1}{a\left(2b+2c-1\right)}+\frac{1}{b\left(2c+2a-1\right)}+\frac{1}{c\left(2a+2b-1\right)}\ge27\)