It is given that a,b,c are positive values. 1) show that (a+b+c)((1/a)+(1/b)+(1/c)) >= 9

2) If a+b+c=1 , show that (2-a),(2-b),(2-c) are positive

3) Using above results show that [(a/(2-a)) + (b/(2-b)) + (c/(2-c)) ] >= 3/5

I need help with 3rd part

Answers
In the first part you must transform the equation so that it fits you:((a+b+c)/a)+((a+b+c)/b)+((a+b+c)/c)>=91+(b/a)+(c/a)+(a/b)+1+(c/b)+(a/c)+(b/c)+1>=9rearranging(a/b+b/a) + (a/c+c/a) + (b/c+c/b) + 3 >=9and you must develop the following remarkable product: (a-b)^2=a^2-2ab+b^2 if a=b then a^2-2ab+b^2=0=> a^2+b^2=2ab => (a^2+b^2)/ab=2 => a/b+b/a=2the important thing here is that you make the assumption a = b anda=c and b=cWith that assumption you have the problem done
luisxxi
23 April 2019
Hi @luisxxi, thanks for taking your time to help me on this question. But as I've mentioned in my post, I need help on the 3rd part. I was able to manage the first two parts myself.Any help on this is highly appreciates.Many thanks!
ashley760
26 April 2019
Add an answer

Similar questions