Let $x=(a+c)(b+d)$. We have $x=\sum_{cyc} ab \le \sum_{cyc} \frac{a^2+b^2}{2}=4$. Note that by Cauchy-Schwarz, \begin{align*} \sum_{cyc} \frac{a^3}{a+b} &=\sum_{cyc} \left( a^2-\frac{a^2b}{a+b} \right) \\ & \le 4- \frac{\left(\sum_{cyc} ab \right)^2}{\sum_{cyc} b(a+b)} \\ &=4-\frac{x^2}{4+x}=\frac{16+4x-x^2}{4+x}. \end{align*}and by AM-GM, \[ 4abcd \le 4 \cdot \frac{(a+c)^2}{4} \cdot \frac{(b+d)^2}{4} =\frac{x^2}{4}. \]So we only need to prove \[ \frac{16+4x-x^2}{4+x}+\frac{x^2}{4} \le 6. \]It is equivalent to \[ \frac{(x-4)(x^2+4x+8)}{4(x+4)} \le 0, \]which is true since $x \le 4$.

$$\sum \frac{a^2(a+b)-a^2b}{a+b}+4abcd \le 6 \leftrightarrow 2+\sum \frac{a^2b}{a+b}\ge 4abcd \leftrightarrow \sum b^2+ \frac{a^2b}{a+b}\ge 4abcd+2 $$$$\leftrightarrow $$$$\sum b \frac{(a+b)^2-ab}{(a+b)} \ge 4abcd+2 \leftrightarrow 2 +\sum ab \ge \sum \frac{ab^2}{a+b}+4abcd$$. $1)$ $\sum ab= \frac{1}{4}\sum ab \sum a^2\ge 4abcd$ $2)$ From Cauchy-Schwartz $4=\frac{1}{4}(\sum b^2)^2\ge (\sum b^2) \frac{1}{4} \sum ab\ge (\sum b^2)(\sum \frac{a^2b^2}{(a+b)^2})\ge (\sum \frac{ab^2}{a+b})^2 \implies 2\ge \sum \frac{ab^2}{a+b}$

hakN wrote: For any positive reals $a,b,c,d$ that satisfy $a^2 + b^2 + c^2 + d^2 = 4,$ show that $$\frac{a^3}{a+b} + \frac{b^3}{b+c} + \frac{c^3}{c+d} + \frac{d^3}{d+a} + 4abcd \leq 6.$$

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sqing wrote: hakN wrote: For any positive reals $a,b,c,d$ that satisfy $a^2 + b^2 + c^2 + d^2 = 4,$ show that $$\frac{a^3}{a+b} + \frac{b^3}{b+c} + \frac{c^3}{c+d} + \frac{d^3}{d+a} + 4abcd \leq 6.$$ I can't understand why it works

Here is another solution. $$\sum_{cyc}\frac{a^3}{a+b}=\sum_{cyc}\frac{a^3+b^3-b^3}{a+b}=\sum_{cyc}(a^2-ab+b^2-\frac{b^3}{a+b})=8-\sum_{cyc}ab-\sum_{cyc}\frac{b^3}{a+b}$$by AM-GM, $$\sum_{cyc}\frac{b^3}{a+b}+\sum_{cyc}\frac{b(a+b)}{4}\ge \sum_{cyc}b^2=4$$Therefore, it is enough to show that $$\frac{3}{4}\sum_{cyc}ab+1\ge 4abcd$$Since $abcd\le 1$, by AM-GM, we have $\frac{3}{4}\sum_{cyc}ab+1\ge 3\sqrt{abcd}+1\ge 4abcd$ and it ends the proof.

Note that by $\text{AM-GM}$ $$(\sum a^2)(\sum ab) \ge 16abcd \implies \sum ab \ge 4abcd(*)$$by $\text{Cauchy-Schwarz and AM-GM}$ $$\sum \frac{b^3}{a+b} = \sum \frac{b^4}{b^2+ab} \ge \frac{(\sum b^2)^2}{\sum b^2+\sum ab} \ge \frac{(\sum b^2)^2}{2 \sum b^2} \ge 2(**)$$ Thus $$\sum \frac{a^3}{a+b} + 4abcd \le 6 \iff \sum \frac{a^3+b^3}{a+b}- \sum \frac{b^3}{a+b}+4abcd \le 6 \iff 2(\sum a^2)-\sum ab -\sum \frac{b^3}{a+b}+4abcd \le 6 \iff 2+4abcd \le \sum \frac{b^3}{a+b}+\sum ab$$Which is true according to $(*)$ and $(**)$ ... So we are done

\[\sum{\frac{a^3}{a+b}}=\sum{(a^2-ab+\frac{ab^2}{a+b})}=4-\sum{ab}+\sum{\frac{ab^2}{a+b}}\leq 4-4\sqrt{abcd}+\sum{\frac{ab^2}{2\sqrt{ab}}}\leq 4-4abcd+\frac{1}{2}\sum{b\sqrt{ab}}\]\[4-4abcd+\frac{1}{2}\sum{b\sqrt{ab}}\leq 4-4abcd+\frac{1}{2}\sqrt{\sum{b^2}\sum{ab}}\leq 4-4abcd+2=6-4abcd\]Since $4=a^2+b^2+c^2+d^2\geq \frac{((a+c)+(b+d))^2}{4}\geq (a+c)(b+d)=ab+bc+cd+da $ and $4=a^2+b^2+c^2+d^2\geq 4\sqrt{abcd}\implies 1\geq abcd$ as desired.$\blacksquare$

I think #4 is most likely wrong The inequality is flipped? #4 proved $6-4abcd\geq \frac{16}{4+(ab+bc+cd+da)}$ but this does not guarantee $6-4abcd\geq \sum_{cyc}\frac{a^3}{a+b}$ because $\sum_{cyc}\frac{a^3}{a+b}\geq \frac{16}{4+(ab+bc+cd+da)}$