Question

Chứng minh bất đẳng thức:

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\) Với \(a\ge b\ge c>0\)

Answer 1

\(\frac{a}{a+b}\)>=  \(\frac{a}{a+a}\)= \(\frac{1}{2}\)( vì a + a >= a + b vì a >= b ) 

\(\frac{b}{b+c}\) >= \(\frac{b}{b+b}\)= \(\frac{1}{2}\)( vì b + b >= b + c vì b >= c )

\(\frac{c}{c+a}\)>= \(\frac{c}{c+c}\)  = \(\frac{1}{2}\)( vì c + c >= c + a vì c>=0 )

Từ 3 điều này suy ra

\(\frac{a}{a+b}\)+ \(\frac{b}{b+c}\)+ \(\frac{c}{c+a}\)>=  \(\frac{3}{2}\)

Answer 2

dễ dàng c/m (x+y+z)(1/x+1/y+1/z) \(\ge\) 9,dấu "=" khi x=y=z (*)

a/a+b +b/b+c +c/c+a >= 3/2

<=>(a/b+c + 1) + (b/c+a + 1) + (c/a+b + 1) >= 3/2+1+1+1

<=>(a+b+c)/(b+c) + (a+b+c)/(c+a) + (a+b+c)/(a+b) >= 9/2

<=>2(a+b+c)(1/b+c + 1/c+a + 1/a+b) >= 9/2

<=>[(b+c)+(c+a)+(a+b)](1/b+c + 1/c+a + 1/a+b) >= 9/2 (bđt (*))