Very hard inequality: $\frac{a}{\sqrt{a+pb}}+\frac{b}{\sqrt{b+pc}}+\frac{c}{\sqrt{c+pa}} \le k_p \sqrt{a+b+c}.$ Given $p>0$. Find the smallest real number $k_p$ such that the following inequality holds for any non-negative reals $a,b,c$:
$$\frac{a}{\sqrt{a+pb}}+\frac{b}{\sqrt{b+pc}}+\frac{c}{\sqrt{c+pa}} \le k_p \sqrt{a+b+c}.$$
Some particular cases:


*

*$k_p = \sqrt{\frac{3}{p+1}}$ for $0\le p\le \frac{1}{2}$.

*$k_1=\frac{5}{4}$.

*$k_{3/2} = \frac{2\sqrt{6}-3\sqrt{2}}{\sqrt{-1+\sqrt{3}}}+\sqrt{-5+3\sqrt{3}}$.

*$k_2 = \frac{2\sqrt{3}-2}{\sqrt{2\sqrt{3}}}+\sqrt{-1+\frac{2}{\sqrt{3}}}$.

*$k_4 = \frac{17 -\sqrt{33}}{6\sqrt{-1+\sqrt{33}}}+\sqrt{\frac{-5+\sqrt{33}}{12}}$.
 A: Consider the vectors $v = \left(\sqrt a,\sqrt b,\sqrt c\right)$ and $u = \left(\frac{\sqrt a}{\sqrt{a + pb}},\frac{\sqrt b}{\sqrt{b + pc}},\frac{\sqrt c}{\sqrt{c + pa}}\right)$
Applying the scalar product inequality, we get:
$$\frac{a}{\sqrt{a+pb}}+\frac{b}{\sqrt{b+pc}}+\frac{c}{\sqrt{c+pa}} = v\cdot u \leq |v||u| = \sqrt{a+b+c}\cdot \sqrt{\frac{a}{a+pb} + \frac{b}{b+pc} + \frac{c}{c+pa}}$$
The equality holds iff the vectors are parallel, that is, iff $\sqrt{a+pb}=\sqrt{b+pc}=\sqrt{c+pa}$
Therefore you must assume $a+pb=b+pc=c+pa\Rightarrow a=b=c$ and the expression on the right side becomes:
$$RHS=\sqrt{a+b+c}\cdot \sqrt{\frac{a}{a+pa} + \frac{a}{a+pa} + \frac{a}{a+pa}} = \sqrt{a+b+c}\cdot \sqrt{\frac{3}{p + 1}}$$
Which is the result you got for $0\leq p \leq \frac{1}{2}$. Actually, for $p=1$, $\sqrt{\frac{3}{p + 1}}$ verifies the inequality and is smaller than $\frac{5}{4}$, so you might want to review that result. In order to finish the problem, it only remains to prove that
$$\sqrt{\frac{a}{a+pb} + \frac{b}{b+pc} + \frac{c}{c+pa}}\leq \sqrt{\frac{3}{p + 1}}$$
