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\(\frac{2+a}{1+a}+\frac{1-2b}{1+2b}=\frac{\left(2+a\right)\left(1+2b\right)+\left(1-2b\right)\left(1+a\right)}{\left(1+a\right)\left(1+2b\right)}=\frac{2a+2b+3}{\left(1+a\right)\left(1+2b\right)}.\)
Ta có: \(\left(2+2a\right)\left(1+2b\right)\le\frac{\left(2+2a+1+2b\right)^2}{4}=\frac{\left(2a+2b+3\right)^2}{4}\)
\(\Rightarrow\left(1+a\right)\left(1+2b\right)\le\frac{\left(2a+2b+3\right)^2}{8}.\)
\(\Rightarrow\frac{2+a}{1+a}+\frac{1-2b}{1+2b}=\frac{2a+2b+3}{\left(1+a\right) \left(1+2b\right)}\ge\frac{2a+2b+3}{\frac{\left(2a+2b+3\right)^2}{8}}=\frac{8}{2a+2b+3}\ge\frac{8}{2.2+3}=\frac{8}{7}.\)
\(VT=1+\dfrac{1}{1+a}+\dfrac{2}{1+2b}-1=2\left(\dfrac{1}{2+2a}+\dfrac{1}{1+2b}\right)\)
\(VT\ge\dfrac{8}{3+2\left(a+b\right)}\ge\dfrac{8}{3+2.2}=\dfrac{8}{7}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=\dfrac{3}{4}\\b=\dfrac{5}{4}\end{matrix}\right.\)
Ta sẽ chứng minh: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)với x,y > 0.
Thật vậy: \(x+y+z\ge3\sqrt[3]{xyz}\)(bđt Cô -si)
và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\)(bđt Cô -si)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)(Dấu "="\(\Leftrightarrow x=y=z\))
Ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
(Dấu "=" xảy ra khi a = b)
Tương tự ta có:\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)(Dấu "=" xảy ra khi b=c)
\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)(Dấu "=" xảy ra khi c=a)
\(VT=\text{Σ}_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
(Dấu "=" xảy ra khi \(a=b=c=\frac{3}{2}\))
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 \(\Leftrightarrow\hept{\begin{cases}a+b=2\\1+a=\frac{1}{2}+b\end{cases}\Leftrightarrow\hept{\begin{cases}a=\frac{3}{4}\\b=\frac{5}{4}\end{cases}}}\)