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 $0\le p\le \frac{1}{2}$.

  • $k_1=\frac{5}{4}$.

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

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

  • $k_4 = \frac{17 -\sqrt{33}}{6\sqrt{-1+\sqrt{33}}}+\sqrt{\frac{-5+\sqrt{33}}{12}}$.

  • $\begingroup$ Some random ramblings... You want the maximum of (LHS) / $\sqrt{a + b + c}$, symmetry suggests the result is along $a = b = c$ or on the borders (some of the three zero). The inequality is homogeneous (multiply all by a constant, you get the same after simplifying), symmetry as above says $a = b = c = 1$, which gives the result you cite for $0 \le p \le 1/2$, are you sure it isn't valid overall?. Also a substitution $u = b/a$, $v = c / a$ might help. $\endgroup$
    – vonbrand
    Apr 1, 2014 at 12:13
  • 1
    $\begingroup$ From a certain value of $p$, equality no longer holds when $a=b=c=1$. For example, for $p=1$ we have $k_1=\frac{5}{4}$ and equality occurs if and only $(a,b,c)$ is proportional to $(3,1,0)$ or its cyclic permutations. $\endgroup$
    – f10w
    Apr 1, 2014 at 12:25
  • 2
    $\begingroup$ If you are interested, try for $p=1$ first. It is a nice result though hard enough (unlike for $0\le p\le \frac{1}{2}$, which is easy by Cauchy-Schwarz). $\endgroup$
    – f10w
    Apr 1, 2014 at 12:29
  • $\begingroup$ Using calculus, the necessary condition for $(1,1,1)$ to be a local maximizer of $f(a,b,c) = \frac{\mathrm{LHS}}{\sqrt{a+b+c}}$ is $p \le \frac{1}{2} + \frac{3}{10}\sqrt{5}\approx 1.1708$; If $p > \frac{1}{2} + \frac{3}{10}\sqrt{5}$, I guess that the maximum of $f(a,b,c)$ is achieved when $c=0$. So I guess that $k_p = \frac{1}{\sqrt{(px_0+1)(x_0+1)}} + \sqrt{\frac{x_0}{x_0+1}}$ where $x_0$ is the smallest positive real root of $(p^3-4p^2)x^3-(p^2+4p)x^2-(p^2-p+1)x + 1 = 0$. When $p=\frac{5}{4}, \frac{3}{2}, 2, 4$, it is the same as your results. $\endgroup$
    – River Li
    Aug 21, 2019 at 15:59

1 Answer 1


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 $$\sqrt{\frac{a}{a+pb} + \frac{b}{b+pc} + \frac{c}{c+pa}}\leq \sqrt{\frac{3}{p + 1}}$$

  • 5
    $\begingroup$ The last inequality may only hold for $p<2$. For $p\ge 2$, it is obviously flipped by the following $$((a+pb)a + (b+pc)b+(c+pa)c)\times (LHS)^2\ge (a+b+c)^2.$$ $\endgroup$ Dec 2, 2015 at 16:44

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