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ta có \(\frac{2+a}{1+b}+\frac{1-2b}{1+2b}=\frac{1+a+1}{1+a}+\frac{2-\left(1+2b\right)}{1+2b}=\frac{1}{1+a}+\frac{2}{1+2b}\)
sử dụng bất đẳng thức Cauchy-Schwwarz ta có:
\(\frac{1}{1+a}+\frac{2}{1+2b}=\frac{1}{1+a}+\frac{1}{\frac{1}{2}+b}\ge\frac{4}{1+a+\frac{1}{2}+b}\ge\frac{4}{1+\frac{1}{2}+2}=\frac{8}{7}\)do a+b =<2
dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b=2\\1+a=\frac{1}{2}+b\end{cases}\Leftrightarrow\hept{\begin{cases}a=\frac{3}{4}\\b=\frac{5}{4}\end{cases}}}\)
\(VT=1+\dfrac{1}{1+a}+\dfrac{2}{1+2b}-1=2\left(\dfrac{1}{2+2a}+\dfrac{1}{1+2b}\right)\)
\(VT\ge\dfrac{8}{3+2\left(a+b\right)}\ge\dfrac{8}{3+2.2}=\dfrac{8}{7}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=\dfrac{3}{4}\\b=\dfrac{5}{4}\end{matrix}\right.\)
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$\frac{2+a}{1+a}=1+\frac{1}{1+a}$
\(\frac{1-2b}{1+2b}=-1+\frac{2}{1+2b}\)$\frac{1-2b}{1+2b}=-1+\frac{2}{1+2b}$
$\frac{1}{1+a}+\frac{2}{2+2b}=\frac{2}{2+2a}+\frac{2}{2+2b}\ge \frac{8}{4+2\left(a+b\right)}=\frac{8}{7}$
Ta có : \(\frac{a}{2a-1}+\frac{b}{2b-1}=\frac{1}{\frac{2a-1}{a}}+\frac{1}{\frac{2b-1}{b}}\ge\frac{4}{4-\left(\frac{1}{a}+\frac{1}{b}\right)}\)
Do đó cần chứng minh \(\frac{4}{4-\left(\frac{1}{a}+\frac{1}{b}\right)}\ge\frac{4}{1+ab}\)
Điều này tương đương với \(4-\left(\frac{1}{a}+\frac{1}{b}\right)\le1+ab\Leftrightarrow\frac{1}{a}+\frac{1}{b}+ab-3\ge0\)
\(\Leftrightarrow\frac{a+b+a^2b^2-3ab}{ab}\ge0\Leftrightarrow\frac{\left(a^2b^2-2ab+1\right)+\left(a+b-ab-1\right)}{ab}\ge0\)
\(\Leftrightarrow\frac{\left(ab-1\right)^2+\left(a-1\right)\left(b-1\right)}{ab}\ge0\)
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
VT = 2/(2+2a) + 2/(1+2b) >= 2. 4/(2+2a+1+2b) >= 8/7
\(\frac{2+a}{1+a}+\frac{1-2b}{1+2b}=\frac{\left(2+a\right)\left(1+2b\right)+\left(1-2b\right)\left(1+a\right)}{\left(1+a\right)\left(1+2b\right)}=\frac{2a+2b+3}{\left(1+a\right)\left(1+2b\right)}.\)
Ta có: \(\left(2+2a\right)\left(1+2b\right)\le\frac{\left(2+2a+1+2b\right)^2}{4}=\frac{\left(2a+2b+3\right)^2}{4}\)
\(\Rightarrow\left(1+a\right)\left(1+2b\right)\le\frac{\left(2a+2b+3\right)^2}{8}.\)
\(\Rightarrow\frac{2+a}{1+a}+\frac{1-2b}{1+2b}=\frac{2a+2b+3}{\left(1+a\right) \left(1+2b\right)}\ge\frac{2a+2b+3}{\frac{\left(2a+2b+3\right)^2}{8}}=\frac{8}{2a+2b+3}\ge\frac{8}{2.2+3}=\frac{8}{7}.\)