Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+b}}\leq\frac{\sqrt{33}}{4}\sqrt{a+b+c+d}$

Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+d}}+\frac{d}{\sqrt{d+a}}\leq\frac{\sqrt{33}}{2}$$

Dear @Andreas! I tried the following C-S for non-negatives $m$ and $n$: $$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a+b}}\right)^2\leq\sum\limits_{cyc}\frac{a}{ka+mb+c}\sum_{cyc}\frac{a(ka+mb+c)}{a+b}$$ From here I didn't find values of $k$ and $m$ for which $$\sum\limits_{cyc}\frac{a}{ka+mb+c}\sum_{cyc}\frac{a(ka+mb+c)}{a+b}\leq\frac{33(a+b+c+d)}{16}$$ is true.

I used also a similar delirium with AM-GM, but I think not all people will like to see it.

• "I tried AM-GM, C-S and more, but without success". That stereotypical information won't help anybody. – Andreas Dec 8 '16 at 14:07
• C-S using $\left(\sum \sqrt{a} \cdot \sqrt{\frac{a}{a+b}}\right)^2 \leq \sum a\sum\frac{a}{a+b}$ looks like it might work – Kibble Dec 8 '16 at 14:08
• the equality case gives $\frac{\sqrt{32}}{2}$, either this is not tight or this is strange – clark Dec 8 '16 at 14:11
• @Andreas, I tried the following C-S $\left(\sum\limits_{cyc}\frac{a}{\sqrt{a+b}}\right)^2\leq\sum\limits_{cyc}\frac{a}{ka+mb+c}\sum\limits_{cyc}\frac{a(ka+mb+c)}{a+b}$ – Michael Rozenberg Dec 8 '16 at 16:21
• @Kibble Your way gets a wrong inequality. – Michael Rozenberg Dec 8 '16 at 16:23

WLOG, assume that $$d = \min(a,b,c,d)$$. Note that $$x\mapsto \sqrt{x}$$ for $$x\ge 0$$ is concave. We have \begin{align} &\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}}\\ =\ & \frac{a}{a+b}\sqrt{a+b} + \frac{8b}{11(b+c)}\sqrt{\frac{121}{64}(b+c)}\\ \le\ & \Big(\frac{a}{a+b} + \frac{8b}{11(b+c)}\Big) \sqrt{\frac{\frac{a}{a+b}}{\frac{a}{a+b} + \frac{8b}{11(b+c)}}(a+b) + \frac{\frac{8b}{11(b+c)}}{\frac{a}{a+b} + \frac{8b}{11(b+c)}}\frac{121}{64}(b+c) }\\ =\ & \sqrt{\frac{(8a+11b)(19ab+11ca + 8b^2)}{88(a+b)(b+c)}} \end{align} and \begin{align} &\frac{c}{\sqrt{c+d}} + \frac{d}{\sqrt{d+a}}\\ =\ & \frac{c}{c+d}\sqrt{c+d} + \frac{d}{d+a}\sqrt{d+a}\\ \le \ & (\frac{c}{c+d} + \frac{d}{d+a})\sqrt{\frac{\frac{c}{c+d}}{\frac{c}{c+d} + \frac{d}{d+a}}(c+d) + \frac{\frac{d}{d+a}}{\frac{c}{c+d} + \frac{d}{d+a}}(d+a) }\\ =\ & \sqrt{\frac{ca+2cd+d^2}{d+a}}. \end{align} Thus, it suffices to prove that $$\sqrt{\frac{(8a+11b)(19ab+11ca + 8b^2)}{88(a+b)(b+c)}} + \sqrt{\frac{ca+2cd+d^2}{d+a}} \le \frac{\sqrt{33}}{4}\sqrt{a+b+c+d}.$$ Denote $$A = \frac{(8a+11b)(19ab+11ca + 8b^2)}{88(a+b)(b+c)}$$ and $$B = \frac{ca+2cd+d^2}{d+a}$$. Squaring both sides of the inequality, it suffices to prove that $$2\sqrt{AB} \le \frac{33}{16}(a+b+c+d) - A - B.$$ Note that $$\frac{33}{16}(a+b+c+d) - A - B = \frac{1}{176(a+b)(b+c)(d+a)} f(a, b, c, d)$$ where $$f(a,b,c,d)$$ is a polynomial with non-negative coefficients. Thus, it suffices to prove that $$\Big(\frac{33}{16}(a+b+c+d) - A - B\Big)^2 - 4AB \ge 0$$ or $$1024g(a,b,c,d) + h(a,b,c,d)\ge 0$$ where $$h(a,b,c,d)$$ is a polynomial with non-negative coefficients, and $$g(a,b,c,d)$$ is a polynomial given by $$g(a,b,c,d) = 3\, a^6\, b^2 + 21\, a^6\, b\, c + 34\, a^6\, c^2 + 20\, a^5\, b^3 - 66\, a^5\, b^2\, c + 48\, a^5\, b^2\, d - 64\, a^5\, b\, c^2 + 217\, a^5\, b\, c\, d - 53\, a^5\, c^3 + 200\, a^5\, c^2\, d + 53\, a^4\, b^4 - 217\, a^4\, b^3\, c + 210\, a^4\, b^3\, d - 144\, a^4\, b^2\, c^2 + 416\, a^4\, b^2\, c\, d - 14\, a^4\, b^2\, d^2 + 25\, a^4\, b\, c^3 + 199\, a^4\, b\, c^2\, d + 213\, a^4\, b\, c\, d^2 + 34\, a^4\, c^4 - 159\, a^4\, c^3\, d + 242\, a^4\, c^2\, d^2 + 65\, a^3\, b^5 - 58\, a^3\, b^4\, c + 366\, a^3\, b^4\, d + 192\, a^3\, b^3\, c^2 + 25\, a^3\, b^3\, c\, d - 50\, a^3\, b^3\, d^2 + 279\, a^3\, b^2\, c^3 - 406\, a^3\, b^2\, c^2\, d + 401\, a^3\, b^2\, c\, d^2 - 55\, a^3\, b^2\, d^3 + 68\, a^3\, b\, c^4 - 330\, a^3\, b\, c^3\, d + 365\, a^3\, b\, c^2\, d^2 + 25\, a^3\, b\, c\, d^3 + 4\, a^3\, c^4\, d - 162\, a^3\, c^3\, d^2 + 79\, a^3\, c^2\, d^3 + 34\, a^2\, b^6 + 79\, a^2\, b^5\, c + 264\, a^2\, b^5\, d + 242\, a^2\, b^4\, c^2 - 216\, a^2\, b^4\, c\, d + 79\, a^2\, b^4\, d^2 + 200\, a^2\, b^3\, c^3 - 302\, a^2\, b^3\, c^2\, d + 167\, a^2\, b^3\, c\, d^2 - 232\, a^2\, b^3\, d^3 + 34\, a^2\, b^2\, c^4 - 23\, a^2\, b^2\, c^3\, d - 186\, a^2\, b^2\, c^2\, d^2 - 6\, a^2\, b^2\, c\, d^3 + 34\, a^2\, b^2\, d^4 + 8\, a^2\, b\, c^4\, d - 408\, a^2\, b\, c^3\, d^2 + 230\, a^2\, b\, c^2\, d^3 + 68\, a^2\, b\, c\, d^4 + 4\, a^2\, c^3\, d^3 + 34\, a^2\, c^2\, d^4 + 68\, a\, b^6\, d - 26\, a\, b^5\, c\, d + 145\, a\, b^5\, d^2 + 111\, a\, b^4\, c^2\, d - 135\, a\, b^4\, c\, d^2 - 230\, a\, b^4\, d^3 + 148\, a\, b^3\, c^3\, d - 414\, a\, b^3\, c^2\, d^2 + 53\, a\, b^3\, c\, d^3 + 6\, a\, b^3\, d^4 + 4\, a\, b^2\, c^4\, d - 238\, a\, b^2\, c^3\, d^2 + 291\, a\, b^2\, c^2\, d^3 + 136\, a\, b^2\, c\, d^4 + 8\, a\, b\, c^3\, d^3 + 68\, a\, b\, c^2\, d^4 + 34\, b^6\, d^2 - 106\, b^5\, c\, d^2 - 53\, b^5\, d^3 - 102\, b^4\, c^2\, d^2 + 83\, b^4\, c\, d^3 + 34\, b^4\, d^4 + 8\, b^3\, c^3\, d^2 + 140\, b^3\, c^2\, d^3 + 68\, b^3\, c\, d^4 + 4\, b^2\, c^3\, d^3 + 34\, b^2\, c^2\, d^4.$$ It suffices to prove that $$g(a,b,c,d)\ge 0$$ for $$d = \min(a,b,c,d)$$. I verify it numerically. Maybe someone can give nice proofs of it.