Finding $\small{\max\limits_{ab+bc+ca=1}\dfrac{\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}}{\sqrt{3a+3b+3c+10}} }$

Question

Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Find the maximal value $P$ $$P=\frac{\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}}{\sqrt{3a+3b+3c+10}} .$$

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When I denote $a=b=1;c=0$ we get $P\le \dfrac{3}{4}.$ We'll prove that is maximal value.

Now, it's enough to prove $$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\le \frac{3}{4} \sqrt{3a+3b+3c+10}.$$ By Cauchy-Schwarz inequality$$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\le \sqrt{3(a+b+c+1)}\le \frac{3}{4} \sqrt{3a+3b+3c+10},$$which is true at $\sqrt{3}\le a+b+c\le 2.$

In remain case, $a+b+c>2.$ It suffices to prove $$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\le 3.$$ I still failed to prove it is true. Hope you help me continue my idea.

Also, does MV (Mixing variables) technique, AM-GM help? Thank you.

Update 1: RiverLi's Cauchy-Schwarz using is very nice but it's still a complicated solution.

Answer 1

Proof. It suffices to prove that $$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\le \frac{3}{4} \sqrt{3a+3b+3c+10}$$ or $$\left(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\right)^2 \le \frac{9}{16} (3a+3b+3c+10).\tag{1}$$

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &\left(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\right)^2\\[6pt] \le{}& \left(\sum_{\mathrm{cyc}}\frac{a + bc}{23ab + 44bc + 23ca + 23a + 2b + 2c}\right)\\[6pt] &\qquad \cdot \left(\sum_{\mathrm{cyc}} (23ab + 44bc + 23ca + 23a + 2b + 2c)\right)\\[6pt] ={}& \left(\sum_{\mathrm{cyc}}\frac{a + bc}{23ab + 44bc + 23ca + 23a + 2b + 2c}\right) \cdot 9(3a + 3b + 3c + 10). \tag{2} \end{align*}

From (1) and (2), it suffices to prove that $$\sum_{\mathrm{cyc}}\frac{a + bc}{23ab + 44bc + 23ca + 23a + 2b + 2c} \le \frac{1}{16}. \tag{3}$$

We use pqr method. Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$. Using $p^2\ge 3q = 3$, we have $p \ge \sqrt 3$.

(3) is equivalently written as $$-1701\,{r}^{2}+ \left( -1767\,{p}^{2}+2841\,p+3978 \right) r + 2(p-2)(2p^2+25p+86) \ge 0.$$ Using $q^2 \ge 3pr$, it suffices to prove that $$-1701\,{r}\cdot \frac{1}{3p}+ \left( -1767\,{p}^{2}+2841\,p+3978 \right) r + 2(p-2)(2p^2+25p+86) \ge 0$$ or $$\left(-\frac{567}{p} -1767\,{p}^{2}+2841\,p+3978 \right) r + 2(p-2)(2p^2+25p+86) \ge 0. \tag{4}$$

We split into two cases.

Case 1. $\sqrt 3 \le p \le 2$

Using $r \ge \frac{4pq - p^3}{9} = \frac{p(4 - p^2)}{9}$ (degree three Schur), using $-\frac{567}{p} -1767\,{p}^{2}+2841\,p+3978 > 0$, it suffices to prove that $$\left(-\frac{567}{p} -1767\,{p}^{2}+2841\,p+3978 \right) \cdot \frac{p(4 - p^2)}{9} + 2(p-2)(2p^2+25p+86) \ge 0$$ or $$\frac13(p - 2)(589p^4 + 231p^3 - 3208p^2 - 2313p + 894) \ge 0$$ which is true.

Case 2. $p > 2$

Clearly, we only need to prove the case that $-\frac{567}{p} -1767\,{p}^{2}+2841\,p+3978 < 0$.

From $0 \le (a-b)^2(b-c)^2(c-a)^2 = -4\,{p}^{3}r+{p}^{2}{q}^{2}+18\,pqr-4\,{q}^{3}-27\,{r}^{2}$, we have $$r \le - \frac{2}{27} p^3 + \frac{p}{3} + \frac{2}{27}(p^2 - 3)\sqrt{p^2 - 3} =: r_2.$$

It suffices to prove that $$\left(-\frac{567}{p} -1767\,{p}^{2}+2841\,p+3978 \right) \cdot r_2 + 2(p-2)(2p^2+25p+86) \ge 0 \tag{5}$$ which is true.

We are done.

Answer 2

Some thought.

I hope the following approach will help.

Assume that $a\ge b\ge c\ge 0.$ Chose $t>0$ such that $t^2+2tc=1.$

It implies that $a\ge t\ge b\ge c$. Now, denote $$\color{black}{f(a,b,c)=\frac{\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}}{\sqrt{3a+3b+3c+10}}.}$$ And $$\color{black}{f(t,t,c)=\frac{2\sqrt{t+tc}+\sqrt{c+t^2}}{\sqrt{6t+3c+10}}.}$$ If we can prove $f(a,b,c)\le f(t,t,c)$ the rest seems smooth by proving $$f(t,t,c)=\frac{2\sqrt{t+t\dfrac{1-t^2}{2t}}+\sqrt{\dfrac{1-t^2}{2t}+t^2}}{\sqrt{6t+3\dfrac{1-t^2}{2t}+10}}\le \frac{3}{4}.$$