Any idea?

Hello. At first,observe that $(a-1)^2\geq 0\Leftrightarrow a^2-2a+1\geq 0\Leftrightarrow a^2+6a+9\geq 8a+8\Leftrightarrow (a+3)^2\geq 8(a+1)\Leftrightarrow $ $a+3\geq 2\sqrt{2(a+1)}\Leftrightarrow \frac{1}{a+3}\leq \frac{1}{\sqrt{8(a+1)}}\Leftrightarrow \frac{8}{a+3}\leq \frac{2\sqrt{2}}{\sqrt{a+1}} \ (1)$. Also,note that $a^2+b^2\geq \frac{(a+b)^2}{2}$ and $4ab\leq (a+b)^2$.Hence, $\frac{8}{(a+b)^2+4abc}+\frac{a^2+b^2}{2}\geq \frac{8}{(a+b)^2+c(a+b)^2}+\frac{(a+b)^2}{4}=\frac{8}{(c+1)(a+b)^2}+\frac{(a+b)^2}{4}\geq \frac{2\sqrt{2}}{\sqrt{a+1}}\overset{(1)}\geq \frac{8}{a+3}$. Repeating this procedure cyclically and adding by parts gives us the desired result.

This problem was proposed by Bosnia.

Let $f(a,\ b,\ c):= \frac{8}{(b+c)^2+4bca}+\frac{b^2+c^2}{2}-\frac{8}{a+3}$ $\geq \frac{8}{(b+c)^2(a+1)}+\left(\frac{b+c}{2}\right)^2-\frac{8}{a+3}$ $=\frac{2}{a+1}\cdot\frac{1}{\left(\frac{b+c}{2}\right)^2}+\left(\frac{b+c}{2}\right)^2-\frac{8}{a+3}$ $\geq \frac{2\sqrt{2}}{\sqrt{a+1}}-\frac{8}{a+3}$ $=\frac{2\sqrt{2}(a+3-2\sqrt{2}\sqrt{a+1})}{(a+3)\sqrt{a+1}}$ $=\frac{2\sqrt{2}}{(a+3)\sqrt{a+1}}(\sqrt{a+1}-\sqrt{2})^2\geq 0.$ $\therefore f(a,\ b, \ c)+f(b,\ c,\ a)+f(c,\ a,\ b)\geq 0.\ Q.E.D.$ The equality holds when $a=b=c=1.$

Let $f(a,\ b,\ c):= \frac{8}{(b+c)^2+4bca}+\frac{b^2+c^2}{2}-\frac{8}{a+3}$ $4(a+3)\{(b+c)^2+4abc\}f(a,\ b,\ c)$ $=32(a+3)+\{2(a+3)(b^2+c^2)-32\}\{(b+c)^2+4abc\}$ $\geq 32(a+3)+\{(a+3)(b+c)^2-32\}\{(b+c)^2+4abc\}$ $\geq 32(a+3)+4\{(a+3)bc-8\}\cdot 4bc(a+1)$ $=16\{(a+1)(a+3)(bc)^2-8(a+1)bc+2(a+3)\}$ $=16\left\{(a+1)(a+3)\left(bc-\frac{4}{a+3}\right)^2+\frac{2}{a+3}(a-1)^2\right\}\geq 0.$ Equality holds when $bc=\frac{4}{a+3}$ and $a=1$, similarly $ca=\frac{4}{b+3}$ and $b=1$ and $ab=\frac{4}{c+3}$ and $c=1$, yielding $f(a,\ b,\ c)+f(b,\ c,\ a)+f(c,\ a,\ b)\geq 0$, Equality : $a=b=c=1.$ Q.E.D.

Is there another approach for solving the problem?

EDIT: Just realized this one is a lot like gravilos's solution .

quangminhltv99 wrote:

$LHS\geq\sum\frac{8}{(a+b)^2+c(a+b)^2}+\sum\frac{a^2+b^2}{2}$ $\geq\sum\left(\frac{8}{(c+1)(a+b)^2}+\frac{(a+b)^2}{4}\right)\geq\sum\frac{2\sqrt{2}}{\sqrt{a+1}}.$ Now, we prove that $\frac{2\sqrt{2}}{\sqrt{a+1}}\geq\frac{8}{a+3}$, which is equivalent to $(a-1)^2\geq 0$. Very nice.

$\sum \frac {1} {(a + b)^ 2 + 4abc} + \frac {a ^ 2 + b ^ 2 + c ^ 2} {8} =\sum \left(\frac {1} {(a + b) ^ 2 + 4abc} + \frac {a ^ 2 + b ^ 2} {16} \right)$ $\geq \sum \left(\frac {1} {2 (a ^ 2 + b ^ 2) + 2c (a ^ 2 + b ^ 2)} + \frac {a ^ 2 + b ^ 2} {16}\right) = \sum \left(\frac {1} {2 (a ^ 2 + b ^ 2) (1 + c)} + \frac {a ^ 2 + b ^ 2} {16}\right) $ $\geq 2 \sum \sqrt {\frac {1} {2 (a ^ 2 + b ^ 2) (1 + c)} \frac {a^2 + b^ 2}{16}} = \sum\frac { 1} {2 \sqrt {2 (1 + c)}} \geq \sum\frac {1} {2 + 1 + c}= \sum\frac {1} {a + 3}.$ Writing the corresponding inequalities for the other pairs and adding shows the asking.

For storage Obviously, from $AM-GM$ inequality we have that $a^2+b^2 \geq 2ab$ and multiply it by $2c$ to get $2c(a^2+b^2) \geq 4abc$ and from $(a-b)^2 \geq 0$ we have that $2(a^2+b^2) \geq (a+b)^2$, so: $(a+b)^2+4abc \leq 2(a^2+b^2)(c+1)$ and $\frac{8}{(a+b)^2+4abc} \geq \frac{8}{2(a^2+b^2)(c+1)}=\frac{4}{(a^2+b^2)(c+1)}$ From $AM-GM$ we have that $\frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2} \geq 2\sqrt{\frac{2}{c+1}}=\frac{4}{\sqrt{2(c+1)}}$ From $AM-GM$ we have that $\frac{c+3}{8}=\frac{c+1+2}{8} \geq \frac{\sqrt{2(c+1)}}{4}$, so $\frac{8}{c+3} \leq \frac{4}{\sqrt{2(c+1)}}$, so from previous we have that $\frac{4}{(a^2+b^2)(c+1)} +\frac{a^2+b^2}{2} \geq \frac{8}{c+3}$ and using previous to get $\frac{8}{(a+b)^2+4abc} +\frac{a^2+b^2}{2} \geq \frac{8}{c+3}$, but also $\frac{8}{(a+c)^2+4abc} +\frac{a^2+c^2}{2} \geq \frac{8}{b+3}$ and $\frac{8}{(b+c)^2+4abc} +\frac{b^2+c^2}{2} \geq \frac{8}{a+3}$. Adding up these inequalities gives us final solution

I think the following solution is new: Note that $a^{2}+\frac{3a+1}{a+3}=\frac{(a+1)^{3}}{a+3}$. Thus, by adding $\sum \frac{3a+1}{a+3}$ to both hands, the inequality can be rewritten as: $8\sum \frac{1}{(a+b)^{2}+4abc}+\sum \frac{(a+1)^{3}}{a+3}\geq 9$. But one can easily obtain that: $\sum \frac{8}{(a+b)^{2}+4abc}\geq \sum \frac{8}{(a+b)^{2}(1+c)}=8\sum \frac{\frac{1}{(a+b)^{2}}}{1+c}$. By applying twice Titu's Lemma: $8\sum \frac{\frac{1}{(a+b)^{2}}^{}}{1+c}\geq 8 \frac{(\sum \frac{1}{a+b})^{2}}{3+a+b+c}\geq 8\sum \frac{(\frac{9}{2(a+b+c)})^{2}}{3+a+b+c}$ On the other hand, by generalized Titu's inequality: $\sum \frac{(a+1)^{3}}{a+3}\geq \frac{(\sum a +3)^{^{3}}}{3(\sum a+9)}$ Thus by letting $x= \sum a$ it suffices to prove that $\frac{162}{x^{2}(x+3)}+\frac{(x+3)^{2}}{3(x+3)}\geq 9$ which is equivalent with:$(x-3)^{2}(x^{4}+18x^{3}+126x^{2}+378x+486)\geq 0$, which is true. Actually, this is not the best solution, but it illustrates that the initial inequality is very weak, as I have applied 4 inequalities to prove it. Nevertheless, I think that this is one of the best inequalities that have appeared in JBMO so far. Congrats to the proposer!

Let $a,b,c $ be positive real numbers.Prove that $$\frac{8}{(a+b)^2 + 4abc} + \frac{8}{(b+c)^2 + 4abc} + \frac{8}{(a+c)^2 + 4abc}+ab+bc+ca\ge \frac{8}{a+3} + \frac{8}{b+3} + \frac{8}{c+3}$$