A four variable inequality Let a,b,c,d be non-negative real numbers satisfying a+b+c+d=3. Show a/(1+2b^3)+b/(1+2c^3)+c/(1+2d^3)+d/(1+2a^3) >= (a^2+b^2+c^2+d^2)/(3). I got this inequality from an IMO preparation club. I no longer do contest math but still I find this interesting. There is a weird equality case : (2,1,0,0) works (whereas (2,0,1,0) does not).

aphathalo 2022-10-02 Answered
A four variable inequality
Let a , b , c , d be non-negative real numbers satisfying a + b + c + d = 3. Show
a 1 + 2 b 3 + b 1 + 2 c 3 + c 1 + 2 d 3 + d 1 + 2 a 3 a 2 + b 2 + c 2 + d 2 3 .
I got this inequality from an IMO preparation club. I no longer do contest math but still I find this interesting. There is a weird equality case : ( 2 , 1 , 0 , 0 ) works (whereas ( 2 , 0 , 1 , 0 ) does not).
So far I've tried regrouping terms to disminish the number of variables or the number of non-zero variables. I've shown that the inequality is implied by the three-variable version
a 1 + 2 b 3 + b 1 + 2 c 3 + c 1 + 2 a 3 a 2 + b 2 + c 2 3
if a + b + c = 3. The three variable version itself is implied by the two variable version, but the two variable version is false.
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Answers (1)

Xavier Jennings
Answered 2022-10-03 Author has 9 answers
By AM-GM
c y c a 1 + 2 b 3 = 3 + c y c ( a 1 + 2 b 3 a ) = 3 c y c 2 a b 3 1 + 2 b 3
3 c y c 2 a b 3 3 b 2 = ( a + b + c + d ) 2 2 c y c a b 3 =
= a 2 + b 2 + c 2 + d 2 + a c + b d 3 a 2 + b 2 + c 2 + d 2 3 .
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