[不等式] 貌似很难的。

Let$a,b,c\geq 0 ,a+b+c=3$ prove that:
\[ \frac{ab}{b^{3}+1}+ \frac{bc}{c^{3}+1}+ \frac{ca}{a^{3}+1}\leq \frac{3}{2}\]

1# pxchg1200


First,we build a lemma:
lemma:
When $x \in (0,3) $ we have:
\[ \frac{36}{x^{3}+1}\leq 16x^{2}-59x+61 \]
proof of the lemma:
\[ \Leftrightarrow (16x^{3}-27x^{2}-9x+25)(x-1)^{2}\geq 0 \]
We can easy see that it true,when $x \in (0,3) $.
Then,Let us back to the original inequality
\[ \frac{ab}{b^{3}+1}+\frac{bc}{c^{3}+1}+\frac{ca}{a^{3}+1}\leq \frac{3}{2}\]
Using the lemma:
we get \[ \sum{ab(16b^{2}-59b+61)}\leq 54 \]
and use our condition $a+b+c=3$ to Homogeneous. after many compute it gives:
\[ 6\sum{a^{4}}-4\sum{a^{3}c}-37\sum{a^{3}b}+91\sum{a^{2}b^{2}}-56\sum{a^{2}bc}\geq 0 \]
Here,we can use Vo Quoc Ba Can's sum of square skills.we only need to check the following things
\begin{equation}
m>0
\end{equation}
\begin{equation}
3m(m+n)\geq p^{2}+pg+g^{2}
\end{equation}
When $ m=6, n=91 ,p=-4,g=-37 $
we can find that $ 3m(m+n)=1746>1533=p^{2}+pg+g^{2} $
Hence we are done!
\blacksquare

有点暴力
……

3# kuing


没办法了，那个lemma还是Can说的。另外的证明就是Mix variable了。那个更暴力。