How prove this inequality $a+b+c+d=4$

Question

let $a,b,c,d$ be postive numbers,and such $a+b+c+d=4$, show that $$\dfrac{a}{\sqrt{a+3b}}+\dfrac{b}{\sqrt{b+3c}}+\dfrac{c}{\sqrt{c+3d}}+\dfrac{d}{\sqrt{d+3a}}\le\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+c^2}+\dfrac{c^2}{c+d^2}+\dfrac{d^2}{d+a^2}$$

Answer 1

Remark: Thank @ablmf for pointing out my mistake. The reason of the mistake is because the inequality in the link below has opposite direction. Here I give another proof. The proof relies on Computer Algebra Systems (CAS).
Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$

Proof: Let $x = \frac{a+3b}{4}, y = \frac{b+3c}{4}, z = \frac{c+3d}{4}, w = \frac{d+3a}{4}$. The inequality is written as \begin{align} &\frac{2a}{a+3b}\sqrt{x} + \frac{2b}{b+3c}\sqrt{y} + \frac{2c}{c+3d}\sqrt{z} + \frac{2d}{d+3a}\sqrt{w}\\ \le \ & \frac{a^2}{a+b^2} + \frac{b^2}{b+c^2} + \frac{c^2}{c+d^2} + \frac{d^2}{d+a^2}. \end{align}

We will use the following bound: $$\sqrt{u} \le \frac{u^2+6u+1}{4u+4}, \quad \forall u \ge 0$$ since $\left(\frac{u^2+6u+1}{4u+4}\right)^2 - u = \frac{(u-1)^4}{16(u+1)^2}$.

With the bound above, it suffices to prove that \begin{align} &\frac{2a}{a+3b}\frac{x^2+6x+1}{4x+4} + \frac{2b}{b+3c}\frac{y^2+6y+1}{4y+4} + \frac{2c}{c+3d}\frac{z^2+6z+1}{4z+4} \\ &\qquad + \frac{2d}{d+3a}\frac{w^2+6w+1}{4w+4}\\ \le\ & \frac{a^2}{a+b^2} + \frac{b^2}{b+c^2} + \frac{c^2}{c+d^2} + \frac{d^2}{d+a^2}. \end{align} Homogenization: Let $Q = \frac{a+b+c+d}{4}$. It suffices to prove that \begin{align} &\frac{2a}{a+3b}\frac{x^2+6xQ+Q^2}{4x+4Q} + \frac{2b}{b+3c}\frac{y^2+6yQ+Q^2}{4y+4Q} + \frac{2c}{c+3d}\frac{z^2+6zQ+Q^2}{4z+4Q} \\ &\qquad + \frac{2d}{d+3a}\frac{w^2+6wQ+Q^2}{4w+4Q}\\ \le\ & \frac{a^2Q}{aQ+b^2} + \frac{b^2Q}{bQ+c^2} + \frac{c^2Q}{cQ+d^2} + \frac{d^2Q}{dQ+a^2}. \end{align} After clearing the denominators, it suffices to prove that $g(a, b, c, d)\ge 0$ where $g(a, b, c, d)$ is a homogeneous polynomial of degree $16$.

The Buffalo Way works. There are several cases to deal with:

1) If $d \le c \le b\le a$: Let $d = 1, c = 1+s, b = 1+s+t, a = 1+s+t+r$ for $s, t, r\ge 0$. Then $g(1+s+t+r, 1+s+t, 1+s, 1)$ is a polynomial in $s, t, r$ with non-negative coefficients. True.

2) If $d \le c \le a \le b$: Let $d=1, c = 1+s, a = 1+s+t, b = 1+s+t+r$ for $s, t, r\ge 0$. We have \begin{align} &g(1+s+t, 1+s+t+r, 1+s, 1)\\ =\ & 2885681152r^2-268435456rs+2885681152s^2 + g(s, t, r) \end{align} where $g(s,t ,r)$ is a polynomial with non-negative coefficients. It is easy to prove that $2885681152r^2-268435456rs+2885681152s^2\ge 0$. True.

The other cases are similar. Omitted. We are done.