Question

[Nguyễn Thị Bình Yên]

2. Cho a, b > 0. CM: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Áp dụng CM các bđt sau:

a)Cho a, b, c > 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4.\) CM:\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le1\)

b)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{a+b=c}{2}\left(a,b,c>0\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

a/ \(VT=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow VT\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\frac{3}{4}\)

b/ \(VT\le\frac{ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{bc}{4}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{ca}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\)

\(VT\le\frac{a}{4}+\frac{b}{4}+\frac{b}{4}+\frac{c}{4}+\frac{c}{4}+\frac{a}{4}=\frac{a+b+c}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)