Let a,b,c,d>0 with $ab+bc+cd+da \le 8$. Prove that:
$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \le \frac{1}{abcd}$
Ineq 3
- Souvik saha
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Re: Ineq 3
By Trivial inequality, we have
\[(a-b)^4 = a^4 -4a^3b + 6a^2 b^2 - 4a b^3 + b^4 \ge 0 \]
\[a^4 + 4a^3 b + 6a^2 b^2+4ab^3 + b^4 \ge 8a^3b + 8ab^3 \]
\[(a+b)^4 \ge 8ab(a^2 + b^2)\]
so
\[\frac{a^2 + b^2}{(a+b)^4} \le \frac{1}{8ab}\]
\[(a-b)^4 = a^4 -4a^3b + 6a^2 b^2 - 4a b^3 + b^4 \ge 0 \]
\[a^4 + 4a^3 b + 6a^2 b^2+4ab^3 + b^4 \ge 8a^3b + 8ab^3 \]
\[(a+b)^4 \ge 8ab(a^2 + b^2)\]
so
\[\frac{a^2 + b^2}{(a+b)^4} \le \frac{1}{8ab}\]
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