Let a,b,c>0 be three real numbers. Prove that:
$\sum\limits_{cyc} \frac{a(b+c)}{\sqrt{(a^2+b^2)(a^2+c^2)}} \le 3$
Ineq
Re: Ineq
QM-AM inequality implies
\[\sum\limits_{cyc} \frac{a(b+c)}{\sqrt{(a^2+b^2)(a^2+c^2)}} \le \sum\limits_{cyc} \frac{2a(b+c)}{{(a+b)(a+c)}} \]
So we have to prove \[\sum\limits_{cyc} \frac{2a(b+c)}{{(a+b)(a+c)}} \le 3\]
Or \[\sum\limits_{cyc} \frac{2a(b+c)}{{(a+b)(a+c)}} = \frac{\sum\limits_{cyc} 2a(b+c)^2}{(a+b)(b+c)(c+a)} \le 3\]
\[\Leftrightarrow \sum\limits_{cyc} 2a(b+c)^2 \le 3(a+b)(b+c)(c+a) \]
Expanding, we get this is the same as proving $8abc \le (a+b)(b+c)(c+a)$, which can be proved using AM-GM on $(a+b),(b+c),(c+a)$.
\[\sum\limits_{cyc} \frac{a(b+c)}{\sqrt{(a^2+b^2)(a^2+c^2)}} \le \sum\limits_{cyc} \frac{2a(b+c)}{{(a+b)(a+c)}} \]
So we have to prove \[\sum\limits_{cyc} \frac{2a(b+c)}{{(a+b)(a+c)}} \le 3\]
Or \[\sum\limits_{cyc} \frac{2a(b+c)}{{(a+b)(a+c)}} = \frac{\sum\limits_{cyc} 2a(b+c)^2}{(a+b)(b+c)(c+a)} \le 3\]
\[\Leftrightarrow \sum\limits_{cyc} 2a(b+c)^2 \le 3(a+b)(b+c)(c+a) \]
Expanding, we get this is the same as proving $8abc \le (a+b)(b+c)(c+a)$, which can be proved using AM-GM on $(a+b),(b+c),(c+a)$.
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Nur Muhammad Shafiullah | Mahi