Question

Cho a, b, c>0. Chứng minh rằng 

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

b. \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

c. \(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le abc\)

Answer 1

a) \(\left(a+b\right)^2\ge4ab\)

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

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\left(đpcm\right)\)

Answer 2

Áp dụng BĐT Cô -si cho 3 số dương:

\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

Answer 3

vì sao \(\left(a+b\right)^2\ge4ab\)

Answer 4

\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

Hiểu chưa didudsui

Answer 5

thank you clk_new