# Inequality: $\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$

If $ab+bc+cd+da\leq 8$ and $a,b,c,d \in \mathbb{R}_{+},$how can I prove the following inequality :

$$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$$

I use : $$\frac{a+b}{2}\geq \frac{2ab}{a+b}$$ or $$\frac{1}{(a+b)^2} \leq \frac{1}{4ab}$$ $$\frac{1}{(a+b)^4} \leq \frac{1}{16a^2b^2}$$ $$\frac{a^2+b^2}{(a+b)^4}\leq \frac{a^2+b^2}{16a^2b^2}$$ but from here I have no idea.

thanks :)

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Multiplying $(ab+bc+cd+da)/8$ to the RHS, it suffices to show that

$$\sum \frac{(a^2+b^2)}{(a+b)^4}\le \sum\frac{1}{8ab}.$$

Very luckily we have

$$\frac{(a^2+b^2)}{(a+b)^4} \le \frac{1}{8ab},$$

which is equivalent to $(a+b)^4 \ge 8ab(a^2+b^2)$ or $(a-b)^4 \ge 0$. Q.E.D.

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that is not a hard question !

$\frac{a^2+b^2}{(a+b)^4}\leq \frac{a^2+b^2}{16a^2b^2}$$\Longrightarrow \frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4}$$=\frac{1}{8}$$(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}+\frac{1}{ad})=\frac{1}{8}\cdot{\frac{ab+bc+cd+ad}{abcd}}$

$\leq \frac{1}{abcd}$

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