strong inquality?

Question

Suppose $a,b,c$ are positive real number. Prove that $$\dfrac{a(b+c)}{(a+b)(a+c)}+\dfrac{b(c+a)}{(b+a)(b+c)}+\dfrac{c(a+b)}{(c+a)(c+b)}\ge \dfrac{(a+b+c)^2}{2(a^2+b^2+c^2)}.$$

I 'd tried to use Cauchy-Schwarz's : $$LHS=\sum_{\{a,b,c\}}\dfrac{(a(b+c))^2}{a(a+b)(a+c)(b+c)}\ge \dfrac{4(ab+bc+ca)^2}{(a+b+c)(a+b)(b+c)(c+a)}.$$ After that, I used $p,q,r$ tranformation and need to prove $$\dfrac{4q^2}{p(pq-r)}\ge \dfrac{p^2}{2(p^2-2q)}$$ I wonder if it is a true inequality? Somebody help me? Or give me some hint?

Answer 1

We write the inequality as $$2+\frac{8abc}{(a+b)(b+c)(c+a)} \geqslant \frac{(a+b+c)^{2}}{a^2+b^2+c^2}.$$ But $$(a+b)(b+c)(c+a) \leqslant \frac{8}{9}(a+b+c)(a^2+b^2+c^2).$$ It's remain to prove that $$2+\frac{9abc}{(a+b+c)(a^2+b^2+c^2)} \geqslant \frac{(a+b+c)^{2}}{a^2+b^2+c^2},$$ equivalent to $$a^2+b^2+c^2+\frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$ Which is true. The proof is completed.

Answer 2

As mentioned, the modified inequality isn't always true.

Here's an approach to solve the original inequality:

We use the symmetric-sum notation $ (x, y, z) = \sum_{sym} a^xb^yc^z$ to simplify the algebraic manipulations.

Cross multiplying the denominator, we have

Original inequality $\Leftrightarrow [ (2,1,0) + (1,1,1) ] \times 2 ( 2,0,0) \geq [( 2,0,0) + 2 (1,1,0)] \times [ (2,1,0) + 1/3 (1,1,1) ] $
$ \Leftrightarrow 2(4, 1, 0 ) + 2(3, 2, 0 ) + 6( 3, 1, 1) + 2 (2,2,1)\geq (4,1,0) + 3 (3,2,0) + 3 (3, 1, 1) + 3 (3, 2, 0) + 5 (2, 2, 1) $
$\Leftrightarrow (4,1,0) + 3 (3, 1, 1) \geq (3, 2, 0) + 3 (2, 2, 1) $.

The last inequality is true by Muirhead term wise.