Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$$
This is my attempt.
We have
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}=\frac{a}{a+\sqrt{(a+b+c)a+bc}}+\frac{b}{b+\sqrt{(a+b+c)b+ca}}+\frac{c}{c+\sqrt{(a+b+c)c+ab}}=\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}=1-\frac{\sqrt{(a+b)(a+c)}}{a+\sqrt{(a+b)(a+c)}}+1-\frac{\sqrt{(b+c)(b+a)}}{b+\sqrt{(b+c)(b+a)}}+1-\frac{\sqrt{(c+a)(c+b)}}{c+\sqrt{(c+a)(c+b)}}$$
 A: $$\sum_\text{cyclic} \frac{a}{a+\sqrt{2013a+bc}}=\sum_\text{cyclic} \frac{a}{a+\sqrt{(a+b+c)a+bc}}=\sum_\text{cyclic}\frac{a}{a+\sqrt{(a+b)(a+c)}} $$
Using AM-GM inequlaity,
$$a^2+bc\ge 2a{\sqrt{bc}}\;\;\Longleftrightarrow \;\;\sqrt{(a+b)(a+c)}\ge \sqrt{ab}+\sqrt{ac} $$
Therefore,
$$\sum_\text{cyclic}\frac{a}{a+\sqrt{(a+b)(a+c)}}\le\sum_\text{cyclic}\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\sum_\text{cyclic}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1 $$
A: Using the Cauchy-Schwarz inequality, we have
$$\sqrt{(b+a)(a+c)} \geqslant \sqrt{ab}+\sqrt{ac}.$$
Therefore
$$\frac{a}{a+\sqrt{2013a+bc}} = \frac{a}{a+\sqrt{a(a+b+c)+bc}} = \frac{a}{a+\sqrt{(b+a)(a+c)}}  $$
$$\leqslant \frac{a}{a+\sqrt{ab}+\sqrt{ac}} = \frac{\sqrt {a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$$
So
$$\sum \frac{a}{a+\sqrt{2013a+bc}} \leqslant \sum \frac{\sqrt {a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}  =1.$$
A: We have for $y>1$ and $a\geq b$ and $a\geq c$: $$p=\left(\frac{\left(abc\right)^{y}}{a^{y}+b^{y}+c^{y}}\right)^{\frac{1}{y}}\leq bc$$
So we have with $ap=q$:
$$ f(a)+f(b)+f(c)=\frac{a}{a+\sqrt{2013a+\frac{q}{a}}}+\frac{b}{b+\sqrt{2013b+\frac{q}{b}}}+\frac{c}{c+\sqrt{2013c+\frac{q}{c}}}$$
The function :
$$g(x)=\frac{x}{x+\sqrt{2013x+\frac{a}{x}p}}$$
Is concave for $x\geq 20$ and assumptions of the OP on $a,b,c$ .
So we apply Jensen's inequality to get :
$$f(a)+f(b)+f(c)\leq f(a)+2f\left(\frac{2013-a}{2}\right)$$
Now $y\to\infty$ and we obtain
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{2013-a}{\frac{\left(2013-a\right)}{2}+\sqrt{\frac{2013\left(2013-a\right)}{2}+bc\cdot\frac{2a}{2013-a}}}\leq 1$$
This inequality is not hard.
