Sum of Squares for $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$

Question

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To clarify: I hope to see a (simple) SOS solution similar to the one of the following 3-variable problem, no other assumptions e.g. $a\ge b\ge c\ge d$. Let us see an example. In the link below, by letting $y = \mathrm{mid}(x, y, z)$, two SOS solutions are given. They are not what I want. It is more difficult to get a SOS solution without the assumption $y = \mathrm{mid}(x, y, z)$ for that problem. Prove $2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \le \left(x^2+y^2+z^2+3xyz\right)^2.$

Another example: Mongolian TST 2008, day 2 problem 3, Ji Chen gave a SOS solution with the assumption $x = \min(x, y, z)$. Actually, I gave a SOS solution without any assumption. https://artofproblemsolving.com/community/c6h205316p11219067

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Problem 1. Let $a, b, c, d \ge 0$. Prove that $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$.

There are quite a few solutions (including my solution). Here, I am particularly interested in $\color{blue}{\textrm{(simple) Sum of Squares (SOS) solutions}}$.

Denote $f(a, b, c, d) = \mathrm{LHS} - \mathrm{RHS}$. Then, $f(x^2, y^2, z^2, w^2) \ge 0$ for all real numbers $x, y, z, w$. However, $f(x^2, y^2, z^2, w^2)$ may not be expressed as SOS (polynomial). I found that $(x^2+y^2+z^2+w^2)^2f(x^2, y^2, z^2, w^2)$ may be expressed as SOS, since numerically $(x^2+y^2+z^2+w^2)^2f(x^2, y^2, z^2, w^2) \approx u^\mathsf{T}Q u$ where $Q$ is $185\times 185$ matrix and $u$ is a vector containing monomials in $x, y, z, w$. But $Q$ is quite large, I have not yet proceeded.

Any comments and solutions are welcome.

Relevant information

For 3-variable problem below, there are quite a few SOS solutions.

Problem 2. Let $a, b, c\ge 0$. Prove that $a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$.

My SOS solution is \begin{align} &a^2+b^2+c^2+2abc+1 - 2(ab+bc+ca) \\ =\ & \frac{1}{2(a+b)^2}\Big[(a^2-ac-b^2+bc-a+b)^2 +(a^2-2ab-ac+b^2-bc+a+b)^2\\ &\qquad\qquad\quad + 4ab(a-b)^2 + 4ab(c-1)^2 + 4abc(a+b-2)^2\Big]. \end{align}

FYI, my non-SOS solution for Problem 1:

WLOG, assume that $a\ge b\ge c\ge d$.

If $cd \ge 1$, then $ab \ge 1$ and $(ab-1)(cd-1) \ge 0$, i.e., $ab+cd \le abcd+1$. Also, $bc + da + ac + bd \le \frac{b^2+c^2}{2} + \frac{d^2+a^2}{2} + \frac{a^2+c^2}{2} + \frac{b^2+d^2}{2} = a^2+b^2+c^2+d^2$. Add them up to get the desired result.

If $cd < 1$, let $u = a - \frac{c+d}{cd+1}, \ v = b - \frac{c+d}{cd+1}$. We have \begin{align} \mathrm{LHS} - \mathrm{RHS} &= u^2 + v^2 - (1-cd)uv +\frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\ &\ge 2|uv| - (1-cd) |uv| + \frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\ &\ge \frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\ &\ge \frac{2c^2d^2-c^2d^2-2cd+1}{cd+1}\\ &= \frac{(cd-1)^2}{cd+1}\\ &\ge 0. \end{align} (Q.E.D.)

Answer 1

Since $$1+abcd\geq2\sqrt{abcd}$$ and after replacing $a$ at $a^2$, $b$ at $b^2$, $c$ at $c^2$ and $d$ at $d^2$, we need to prove that $$a^4+b^4+c^4+d^4+2abcd\geq a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2$$ for still non-negative variables, which is true by SOS!

Indeed, let $a\geq b\geq c\geq d$.

Thus, $$a^4+b^4+c^4+d^4+2abcd-(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)=$$ $$=\frac{1}{12}\sum_{sym}(2a^4-3a^2b^2+abcd)=\frac{1}{12}\sum_{cyc}(2a^4-2a^2b^2-a^2b^2+abcd)=$$ $$=\frac{1}{12}\sum_{cyc}((a^2-b^2)^2-a^2b^2+abcd)=$$ $$=\frac{1}{24}\sum_{cyc}(2(a^2-b^2)^2-2a^2b^2+2a^2bc-2a^2bc+2abcd)=$$ $$=\frac{1}{24}\sum_{cyc}(2(a^2-b^2)^2-d^2(a-b)^2-cd(a^2-ab)-cd(b^2-ab))=$$ $$=\frac{1}{48}\sum_{cyc}(4(a^2-b^2)^2-d^2(a-b)^2-c^2(a-b)^2-cd(2a^2-2ab)-cd(2b^2-2ab))=$$ $$=\frac{1}{48}\sum_{sym}(a-b)^2(4(a+b)^2-(c+d)^2)\geq0,$$ where the last inequality is true by the standard SOS's reasoning.

Answer 2

A very nice SOS solution by KaiRain@AoPS:

We have \begin{align*} &a^2+b^2+c^2+d^2+abcd+ 1 - ( ab+bc+cd+da + ac+bd)\\ =\,& \frac{\left(2a^2-2ab+2b^2-ac-ad-bc-bd+2 \right)^2+3 \left(a-b \right)^2 \left(c-d \right)^2 }{4 \left(a^2-ab+b^2+1 \right)}\\ &\quad + \frac{abcd \left(a-b \right)^2+cd \left(ab-1 \right)^2 +\left(c-d \right)^2}{a^2-ab+b^2+1 }. \end{align*}