Ineq 3

For discussing Olympiad Level Algebra (and Inequality) problems
yo79
Posts:53
Joined:Mon Feb 04, 2013 1:01 am
Ineq 3

Unread post by yo79 » Tue Apr 16, 2013 10:29 pm

Let a,b,c,d>0 with $ab+bc+cd+da \le 8$. Prove that:
$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \le \frac{1}{abcd}$

User avatar
Souvik saha
Posts:6
Joined:Sat Apr 13, 2013 12:47 pm
Location:Mymensingh, Dhaka, Bangladesh
Contact:

Re: Ineq 3

Unread post by Souvik saha » Sun Dec 08, 2013 5:44 pm

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
\[\frac{a^2 + b^2}{(a+b)^4} \le \frac{1}{8ab}\]
If you can’t make it good, at least make it look good .Bill Gates

Post Reply