Where did the square roots go?

Since the R.H.S of the inequality is not expressed in a single fraction, this suggests that Titu's Lemma is not a good approach. There seems to be no easy way to simplify the L.H.S. into the desired form on R.H.S. (the square roots are especially cubersome) at first glance.

Since the main method for clearing fractions in inequalities (Cauchy-Schwarz) won't work, wishful thinking compels me to divide and conquer (splitting the initial inequality into $3$ parts before adding them up, hoping that I will somehow end up with the desired form). This idea is motivated by the fact that the symmetry and number of terms of L.H.S. is preserved on the R.H.S. of the inequality.

Therefore, we try to solve $\frac{a}{\sqrt{a+bc}} \geq \frac{Ka}{a+1}$ .

Clearing factors and cross-multiplying, we get $K\sqrt{a+bc} \leq a+1$ .

Since the inequality is symmetric in structure, drawing from previous experience, we have strong reasons to suspect that the equality condition is $a=b=c$ . Motivated by this, we substitute $a=b=c=\frac{1}{3}$ into the inequality and obtain $\frac{2}{3}K \leq \frac{4}{3}$ , so $K \leq 2$ .

Let's prove that the the inequality above is indeed satisfied when $K=2$ . Note that we have

$2\sqrt{a+bc} \leq a+1$

$4(a+bc) \leq a^{2}+2a+1$

$a^{2}-2a+1-4bc \geq 0$

$(a-1)^{2} -4bc \geq 0$

$(-b-c)^{2}-4bc \geq 0$

$(b-c)^{2} \geq 0$ , which is obviously true, with equality holds only when $b=c$ .

Therefore, we have

$\frac{a}{\sqrt{a+bc}} \geq \frac{2a}{a+1}$ .

$\displaystyle \sum_{cyc} \frac{a}{\sqrt{a+bc}} \geq \displaystyle \sum_{cyc} \frac{2a}{a+1}$ .

Equality holds when $a=b=c=\frac{1}{3}$ so the answer is $2+\frac{1}{3} \approx \boxed{2.33}$ .

Let us call the L.H.S of the inequality P

By AM-GM

$a+b+c\ge a+2\sqrt { bc } \Rightarrow \frac { 1-a }{ 2 } \ge \sqrt { bc }$

$bc\le \frac { { \left( 1-a \right) }^{ 2 } }{ 4 }$

Substituting maximum value of bc in the denominator of first term of P

$\frac { a }{ \sqrt { a+bc } } \ge \frac { a }{ \sqrt { a+\frac { { (1-a) }^{ 2 } }{ 4 } } } =\frac { a }{ \sqrt { \frac { { (a+1) }^{ 2 } }{ 4 } } } =\frac { 2a }{ a+1 }$

Doing similarly for the whole expression,

$P\ge \frac { 2a }{ a+1 } +\frac { 2b }{ b+1 } +\frac { 2c }{ c+1 } =2(\frac { a }{ a+1 } +\frac { b }{ b+1 } +\frac { c }{ c+1 })$

Since in AM-GM ,equality occurs when a=b=c= $\frac { 1 }{ 3 }$

Hence K=2 and x=0.33.Thus K+x = $\boxed { 2.33 }$