Problem: Let $a, b, c, d > 0$ with $abcd = 1$. Prove that $$\frac{4}{a^2 + 2b^2 + 3c^2 + 10} \le \frac{5a + 3b + c + 7d}{16(a+b+c+d)}.$$
Background Information: It is verified by Mathematica. This inequality implies the following
Problem 1(Baltic Way 2018): If $a, b, c, d > 0$ with $abcd=1$, then $$\frac{1}{\sqrt{a+2b+3c+10}}+\frac{1}{\sqrt{b+2c+3d+10}}+\frac{1}{\sqrt{c+2d+3a+10}}+\frac{1}{\sqrt{d+2a+3b+10}} \le 1.$$ https://artofproblemsolving.com/community/c6h1734182p11252365
To solve Problem 1, I tried isolated fudging (Why does this Olympiad inequality proving technique (Isolated Fudging) work?), that is: hope to find $p_1, p_2, p_3, r$, such that, for all $a,b,c,d>0; \ abcd=1$, $$ \frac{4}{a+2b+3c+10} \le \frac{p_1a^r+p_2b^r+p_3c^r+(1-p_1-p_2-p_3)d^r}{a^r + b^r+c^r+d^r}. $$ It is not difficult to obtain $p_1 = \frac{5}{16}, \ p_2 = \frac{3}{16}, \ p_3 = \frac{1}{16}$ and $r = 1/2$. Then we get our starting problem. However, I found it is not easy to prove it. By using $a^2 + 2b^2 + 3c^2\ge \frac{1}{6}(a+2b+3c)^2$ (QM-AM), it suffices to prove that $$\frac{4}{\frac{(a+2b+3c)^2}{6}+10} \le \frac{5a+3b+c+7d}{16(a+b+c+d)}$$ which is verified by Mathematica. However, it is still not easy.
Any comments and solutions are welcome and appreciated.
By the way, there is a nice solution to Problem 1: We have \begin{align} &\sum_{\mathrm{cyc}} \frac{4}{a^4+2b^4+3c^4+10}\\ =\ & \sum_{\mathrm{cyc}} \frac{4}{(a^4+b^4+c^4+1) + (b^4+c^4+1+1) + (c^4+1+1+1) + 4}\\ \le\ & \sum_{\mathrm{cyc}} \frac{4}{4abc+4bc+4c+4}\\ =\ & 1 \end{align} which implies Problem 1.
