Bài 1 :
Với x , y > ta chứng minh :
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x-y\right)^2\ge0\) ( luôn đúng )
\(\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng vào bài toán ta có :
\(\frac{1}{a+b+2c}=\frac{1}{a+c+b+c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(\Rightarrow\frac{4ab}{a+b+2c}\le\frac{ab}{a+c}+\frac{ab}{b+c}\)
Tương tự ta cũng có :
\(\frac{4bc}{b+c+2a}\le\frac{bc}{a+b}+\frac{bc}{a+c};\frac{4ca}{c+a+2b}\le\frac{ca}{b+c}+\frac{ca}{a+b}\)
Cộng 3 bất đẳng thức trên vế theo vế ta được :
\(4\left(\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\right)\le\frac{bc+ca}{a+b}+\frac{ab+ca}{b+c}+\frac{ab+bc}{a+c}=c+a+b\)
\(\RightarrowĐpcm\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Bài 2 :
\(Q=\frac{1}{a^2+b^2}+\frac{2102ab+1}{ab}+4ab=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(4ab+\frac{1}{4ab}\right)+\frac{1}{4ab}+2012\)
Áp dụng BĐT : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y};\left(x+y\right)^2\ge4xy\) ta có :
\(\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}=\frac{4}{\left(a+b\right)^2}\ge\frac{4}{1}=4\)
\(\left(4ab+\frac{1}{4ab}\right)^2\ge4.4ab.\frac{1}{4ab}=4\Rightarrow4ab+\frac{1}{4ab}\ge2\)
\(\left(a+b\right)^2\ge4ab\Rightarrow\frac{1}{ab}\ge\frac{4}{\left(a+b\right)^2}\ge\frac{4}{1}=4\Rightarrow\frac{1}{4ab}\ge1\)
\(\Rightarrow Q\ge4+2+1+2012=2019\)
Dấu " = " xay ra khi \(a=b=c=\frac{1}{2}\)
\(\frac{ab}{a+c+b+c}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\) ; \(\frac{bc}{b+c+2a}\le\frac{1}{4}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)\); \(\frac{ca}{c+a+2b}\le\frac{1}{4}\left(\frac{ca}{a+b}+\frac{ca}{b+c}\right)\)
Cộng vế với vế:
\(\Rightarrow VT\le\frac{1}{4}\left(\frac{ab+bc}{a+c}+\frac{ab+ca}{b+c}+\frac{bc+ca}{a+b}\right)=\frac{1}{4}\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
\(1\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le\frac{1}{4}\)
\(Q=\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab+2012\)
\(Q=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{4ab}+4ab+\frac{1}{4ab}+2012\)
\(Q\ge\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{4ab}{4ab}}+\frac{1}{4.\frac{1}{4}}+2012\)
\(Q\ge\frac{4}{1^2}+2+1+2012=2019\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
