$\\$
`a,`
`(a+b+c)(1/a+1/b+1/c)`
`=1+a/b+a/c+b/a+1+b/c+c/a+c/b+1`
`=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)`
Áp dụng BĐT AM-GM ta được :
`a/b+b/a\ge 2\sqrt{a/b . b/a}=2`
`a/c+c/a\ge 2\sqrt{a/c . c/a}=2`
`b/c+c/b\ge 2\sqrt{b/c . c/b}=2`
`-> 3+(a/b+b/c)+(a/c+c/a)+(b/c+c/b)\ge 9`
`->(a+b+c)(1/a+1/b+1/c)\ge 9`
Dấu "`=`" xảy ra khi : `a=b=c`
`b,`
`a/(b+c)+b/(c+a)+c/(a+b)`
`=(a/(b+c)+1)+(b/(c+a)+1)+(c/(a+b)+1)-3`
`=(a+b+c)/(b+c)+(a+b+c)/(c+a)+(a+b+c)/(a+b)-3`
`=2(a+b+c)1/2 (1/(b+c)+1/(c+a)+1/(a+b))-3` (1)
Đặt `b+c=x, c+a=y, a+b+z`
`->x+y+z=2(a+b+c)`
Khi đó (1) trở thành :
`(x+y+z)(1/x+1/y+1/z)1/2-3`
`=(1+x/y+x/z+y/x+1+y/z+z/x+z/y+1)1/2 -3`
`= [3+(x/y+y/x)+(x/z+z/x)+(y/z+z/y)] 1/2 -3`
Áp dụng BĐT AM-GM ta được :
`x/y+y/x\ge 2`
`x/z+z/x\ge 2`
`y/z+z/y\ge 2`
`-> 3 + (x/y+y/x)+(x/z+z/x)+(y/z+z/y)\ge 9`
`-> a/(b+c)+b/(c+a)+c/(a+b) \ge 9 . 1/2-3`
`-> a/(b+c)+b/(c+a)+c/(a+b) \ge 3/2`
Dấu "`=`" xảy ra khi : `a=b=c`
