cho hai số không âm a,b thỏa mãn: \(a+b\le3\)
CMR: \(\frac{2+2a}{1+2a}+\frac{1-4b}{1+4b}\ge\frac{8}{15}\)
chứng minh rằng\(\frac{2+2a}{1+2a}+\frac{1-4b}{1+4b}\ge\frac{8}{5}\) biết \(a+b\le3\)và a, b không âm
Cho a,b,c>0 thỏa mãn a+2b+3c=1
CMR: \(\frac{2ab}{a^2+4b^2}+\frac{6bc}{4b^2+9c^2}+\frac{3ac}{9c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}\right)\ge\frac{15}{4}\)
cho a,b>0 thỏa mãn a+b=4ab. CMR
\(\frac{a}{4b^2+1}+\frac{b}{4a^2+1}\ge\frac{1}{2}\)
vào tcn của tui ấn vào Thông kê hỏi đáp kéo xuống
cho \(a\ge\frac{1}{2},b>1\)Cmr \(\frac{2a^3+1}{4b\left(a-b\right)}=1\)
cho \(a\ge\frac{-1}{2};\frac{a}{b}>1\)cmr \(\frac{2a^3+1}{4b\left(a-3\right)}\ge3\)
Cho 3 số thực dương a, b, c thỏa mãn: \(12\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
CMR: \(\frac{1}{4a+b+c}+\frac{1}{a+4b+c}+\frac{1}{a+b+4c}\le\frac{1}{6}\)
Cho a>0, b<0 thỏa mãn a+b>=0. CMR: \(\frac{1}{a}\ge\frac{2}{b}+\frac{8}{2a-b}\)
Cho a,b là các số thực dương thỏa mãn a + b = 4ab
CMR: \(\frac{a}{4b^2+1}+\frac{b}{4a^2+1}\ge\frac{1}{2}\)
Mong các bạn giúp mình sớm.
\(a+b=4ab\le\left(a+b\right)^2\)
\(\frac{a}{4b^2+1}+\frac{b}{4a^2+1}=\frac{a^2}{4b^2a+a}+\frac{b^2}{4a^2b+b}\)
\(\ge\frac{\left(a+b\right)^2}{4ab\left(a+b\right)+\left(a+b\right)}=\frac{\left(a+b\right)^2}{\left(a+b\right)^2+\left(a+b\right)}\ge\frac{\left(a+b\right)^2}{\left(a+b\right)^2+\left(a+b\right)^2}=\frac{1}{2}\)
\("="\Leftrightarrow a=b=\frac{1}{2}\)
CMR: \(\frac{2a^3+1}{4b\left(a-b\right)}\ge3\) \(\forall\left\{{}\begin{matrix}a\ge\frac{1}{2}\\\frac{a}{b}>1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a>0\\\frac{a}{b}>1\end{matrix}\right.\) \(\Rightarrow b>0\Rightarrow a>b\Rightarrow a-b>0\)
\(\Rightarrow4.b\left(a-b\right)\le\left(b+a-b\right)^2=a^2\)
\(\Rightarrow P=\frac{2a^3+1}{4b\left(a-b\right)}\ge\frac{2a^3+1}{a^2}=2a+\frac{1}{a^2}=a+a+\frac{1}{a^2}\ge3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=1\\b=\frac{1}{2}\end{matrix}\right.\)