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Find the minimum possible value of ![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](//latex.artofproblemsolving.com/4/d/2/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
, 
, 
, 
are nonnegative real numbers such that 
.
Proposed by Titu Andreescu
![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](http://latex.artofproblemsolving.com/4/d/2/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
, 
, 
, 
are nonnegative real numbers such that 
.Proposed by Titu Andreescu
This post has been edited 3 times. Last edited by djmathman, Jun 22, 2020, 5:50 AM
.
from both sides, we get 
But since 
true for 
,
which is true.
and cyclic variants.
![\[ \frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]](http://latex.artofproblemsolving.com/b/2/3/b238c461a3ef0839e20f14f8b6d31054b7bab640.png)
since 
.![\[ ab+bc+cd+da = (a+c)(b+d) \le \left( \frac{(a+c)+(b+d)}{2} \right)^2 = 4. \]](http://latex.artofproblemsolving.com/e/a/0/ea0628772ecacef6a9da5ce9d5bdee5892f83ca9.png)
Thus ![\[ \sum_{\text{cyc}} \frac{a}{b^3+4} \ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \ge 1 - \frac13 = \frac23. \]](http://latex.artofproblemsolving.com/1/f/e/1fe63ef65bd7be5da67cbef0c71c62fda077db39.png)
This minimum 
is achieved at 
and permutations.
is the minimum case...
I also wrote down a short proof for the maximum being 1 cuz I had time left.
then we have 
and 
is positive for 
and is positive for small 
and has only one root between 
and 
so by rolles theorem and stuff 
is nonnegative on all nonnegatives. Thus we have ![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4} \ge \frac{a+b+c+d}4-\frac{ab+bc+cd+da}{12}=1-\frac{(a+c)(b+d)}{12}\ge 1-\frac{4}{12}=\frac 23\]](http://latex.artofproblemsolving.com/d/6/2/d6260e0d7221de323ba0ca9c9d1efe154b1d5e84.png)
by AM-GM with equality when 
.
.
to 
,
,
,
.
and permutations. We claim that ![\[\frac{1}{a^3+4} \geq \frac{-a+3}{12}\]](http://latex.artofproblemsolving.com/d/8/f/d8fadd128f30df168cc3b79e3630b55e75db19cc.png)
for ![$a \in [0, 4]$](http://latex.artofproblemsolving.com/c/2/3/c23f3cce1f1f41978d612789e8ca163584de333b.png)
.
we get ![\[b^4-3b^3+4b \geq 0 \Rightarrow b(b+1)(b-2)^2 \geq 0\]](http://latex.artofproblemsolving.com/3/f/8/3f8705768a1f3c56e997c1840165ae286ab4152f.png)
which is obviously true for ![\[\sum_{\text{cyc}} \frac{3a-ab}{12} \geq \frac{2}{3} \Rightarrow \sum_{\text{cyc}} \frac{-ab}{12} \geq \frac{-1}{3} \Rightarrow \sum_{\text{cyc}} ab \leq 4\]](http://latex.artofproblemsolving.com/6/5/0/6508c9a7baab7fe32708854894f084e7694d7b60.png)
This last inequality is true because ![\[\sum_{\text{cyc}} ab = (a+c)(b+d) = (2-x)(2+x) \leq 4\]](http://latex.artofproblemsolving.com/5/a/9/5a98ba030ed6fc943deb876c10c8927fb196d1db.png)
.![\[\frac{1}{b^3+4} \geq \frac{1}{4} - \frac{b}{12}.\]](http://latex.artofproblemsolving.com/4/1/2/412925a2200b5b8327dd00052ad8601c8785b4e6.png)
![\[(a+c)(b+d) \leq \left(\frac{a+b+c+d}{2}\right)^2 = 4.\]](http://latex.artofproblemsolving.com/9/5/9/9596a00af2884adf224aebdf426f1ea6e6b60130.png)
![\[\sum_{cyc}\frac{a}{b^3+4} \geq \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \geq 1 - \frac{1}{3} = \frac{2}{3}.\]](http://latex.artofproblemsolving.com/d/6/d/d6dee1f2b5c2d7f2cadbfc8c7b6878e03b78e9bd.png)
for nonnegative 
with equality at 
Next note that 
Therefore the original expression is
which is achievable at 
Equality holds when 
from TLT. Notice that expanding gives us 
.
,
.
which is clearly true. Hence, the claim is true. 
we find the LHS is greater than or equal to 
.
.
.
,
This can be seen by expanding, and then we have 
Thus, we have that our expression simplifies to: 
Notice that 
Hence, we have that our answer is 
which occurs at 
and its cyclic permutations.
