Question

Cho a,b,c > 0 

CMR       \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)   \(\ge6\)

Answer 1

\(\frac{a+b}{c}\)+\(\frac{b+c}{a}\)+\(\frac{c+a}{b}\)=\(\frac{a}{c}\)+\(\frac{b}{c}\)+\(\frac{b}{a}\)+\(\frac{c}{a}\)+\(\frac{c}{b}\)+\(\frac{a}{b}\)

Vì a;b;c>0  áp dụng bất đẳng thức cosi ta có:

\(\frac{a}{c}\)+\(\frac{c}{a}\)\(\ge\)2\(\sqrt{\frac{a}{c}.\frac{c}{a}}\)=2

\(\frac{b}{c}\)+\(\frac{c}{b}\)\(\ge\)2\(\sqrt{\frac{b}{c}.\frac{c}{b}}\)=2

\(\frac{b}{a}\)+\(\frac{a}{b}\)\(\ge\)2\(\sqrt{\frac{b}{a}.\frac{a}{b}}\)=2

Cộng vế với vế ta có:

\(\frac{a}{c}\)+\(\frac{b}{c}\)+\(\frac{b}{a}\)+\(\frac{c}{a}\)+\(\frac{c}{b}\)+\(\frac{a}{b}\)\(\ge\)2+2+2

=>\(\frac{a+b}{c}\)+\(\frac{b+c}{a}\)+\(\frac{c+a}{b}\)\(\ge\)6

dấu = xảy ra a=b=c

Answer 2

a=b=c