Proof of an inequality involving 4 algebraic terms

Question

If $a^2+b^2+c^2+d^2 = 4 $ where $a,b,c,d \in R,$ then show that $\frac{a}{b+3} + \frac{b}{c+3} + \frac{c}{d+3} + \frac{d}{a+3} \le 1$.

Answer 1

I have a proof for non-negative variables.

Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.

Hence, by Rearrangement $$\sum\limits_{cyc}\frac{a}{b+3}\leq\frac{x}{t+3}+\frac{y}{z+3}+\frac{z}{y+3}+\frac{t}{x+3}.$$ We'll prove now that $$\frac{x}{t+3}+\frac{t}{x+3}\leq\frac{3x^2+3t^2+10}{32}$$ for all non-negatives $x$ and $t$.

From here we'll get $$\sum_{cyc}\frac{a}{b+3}\leq\frac{3x^2+3t^2+10}{32}+\frac{3y^2+3z^2+10}{32}=1,$$ which ends the proof.

Indeed, let $x+t=2u$ and $xt=v^2$.

Hence, we need to prove that$$\frac{x}{t+3}+\frac{t}{x+3}\leq\frac{3x^2+3t^2+10}{32}$$ or $$3v^4-2(3u^2-9u+5)v^2-(2u+3)(18u^2-32u+15)\leq0.$$

Since $-(2u+3)(18u^2-32u+15)<0$ and $v^2\leq u^2$, it remains to prove that $$3v^4-2(3u^2-9u+5)v^2-(2u+3)(18u^2-32u+15)\leq0$$ for $v^2=u^2$, which gives $(u-1)^2(u^2+8u+15)\geq0$. Done!