Let a,b,c be positive reals such that a+b+c=1. Prove that (bc)/(a+5)+(ca)/(b+5)+(ab)/(c+5)<=1/4. Progress: This is equivalent to show that (1)/(a^2+5a)+(1)/(b^2+5b)+(1)/(c^2+5c)<=(1)/(4abc). I'm not sure how to proceed further. Also equality doesn't occur when a=b=c, which makes me more confused.

Lustyku8 2022-09-20 Answered
Inequality with a+b+c=1
Let a , b , c be positive reals such that a + b + c = 1. Prove that
b c a + 5 + c a b + 5 + a b c + 5 1 4 .
Progress: This is equivalent to show that
1 a 2 + 5 a + 1 b 2 + 5 b + 1 c 2 + 5 c 1 4 a b c .
I'm not sure how to proceed further. Also equality doesn't occur when a = b = c, which makes me more confused.
Edit: The original inequality is as follows: Let a 1 , a 2 , , a n be n > 2 positive reals such that a 1 + a 2 + + a n = 1. Prove that,
k = 1 n j k a j a k + n + 2 1 ( n 1 ) 2 .
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Answers (2)

Miya Swanson
Answered 2022-09-21 Author has 11 answers
Notice that the conditions imply that 0 a , b , c 1; in particular, we have
b c a + 5 + c a b + 5 + a b c + 5 b c 5 + c a 5 + a b 5 b + c + a 5 < 1 4 .
I don't think this works for the general case though, because for n 4, we have ( n 1 ) 2 n + 2
Now for the general case, by AM-GM inequality
j k a j ( j k a j n 1 ) n 1 = ( 1 a k n 1 ) n 1 1 ( n 1 ) n 1 ,
and thus
k = 1 n j k a j a k + n + 2 n ( n + 2 ) ( n 1 ) n 1 1 ( n 1 ) n 1 1 ( n 1 ) 2
since n 1 2
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ct1a2n4k
Answered 2022-09-22 Author has 1 answers
We'll prove a stronger inequality: c y c b c 5 + a 1 16
Indeed, we need to prove that
c y c b c 6 a + 5 b + 5 c a + b + c 16
or
c y c ( a 16 b c 6 a + 5 b + 5 c ) 0
or
c y c 6 a 2 + 5 a b + 5 a c 16 b c 6 a + 5 b + 5 c 0
or
c y c ( a b ) ( 3 a + 8 c ) ( c a ) ( 3 a + 8 b ) 6 a + 5 b + 5 c 0
or
c y c ( a b ) ( 3 a + 8 c 6 a + 5 b + 5 c 3 b + 8 c 6 b + 5 a + 5 c ) 0
or
c y c ( a b ) 2 ( 15 a + 15 b + 7 c ) ( 6 a + 5 b + 5 c ) ( 6 b + 5 a + 5 c ) 0.
Done!
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