<s>[INST] A tough inequality problem with condition $a+b+c+abc=4$ If, $a+b+c+abc=4$, with $a,b,c$ being positive reals, then prove or disprove the following inequality:
$$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geq\frac{a+b+c}{\sqrt2}$$
I couldn't do anything, please help. However, from the some values I have tried, the inequality does seem to be true.
 [/INST] We have by Holder's inequality:
$$\left(\sum_{cyc} \frac{a}{\sqrt{b+c}}\right)^2\left(\sum_{cyc}a(b+c) \right) \ge (a+b+c)^3$$
So it is sufficient to show that
$$2(a+b+c) \ge \sum_{cyc} a(b+c)  \text{ or equivalently, } a+b+c \ge ab + bc + ca$$
Suppose $a+b+c < ab + bc + ca$.  Then by Schur's inequality we have
$$\begin{align}
\frac{9abc}{a+b+c} &\ge 4(ab+bc+ca)- (a+b+c)^2 \\
&> (a+b+c)\left( 4- (a+b+c)\right) \\
&= (a+b+c) \cdot abc \\
\end{align}$$
This gives $a+b+c< 3$, further we have from $4= a+b+c+ abc \ge 4\sqrt{abc} \implies abc \le 1$, so $4 = a+b+c+abc < 3+1$, a contradiction.  Hence proved.
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