Let us consider the function $f(x)=\ln(P+\frac{1}{x})$ Then we have that $f'(x)=\frac{-1}{Px+x^2}$ If $i>j>0$ then $(P+i+j)(i-j)>0$, so $Pi+i^2>Pj+j^2$. Taking the negative reciprocal gets that $\frac{-1}{Pi+i^2}>\frac{-1}{Pj+j^2}$ for all $i>j$ Therefore $f$ is convex on $(0,\infty)$ Therefore, for all $x,y>0$, the statement $\frac{1}{2}(f(x)+f(y))\ge f(\frac{x+y}{2}) $ is true. Plugging in $(a+2b+3c,3a+2b+c),(4a+2b,4b+2a),(4a+2b,4a+2c),$ and $(a+2b+3c,3a+2b+c)$ for $(x,y)$ then summing results in the following: $f(a+2b+3c)+f(3a+2b+2c)+f(4a+2b)+\frac{1}{2}(f(4a+2c)+f(4b+2a))\ge 2f(2(a+b+c))+f(3a+3b)+f(4a+b+c)$ subtracting $f(3a+2b+2c)+f(a+2b+3c)+2f(2(a+b+c))$ from both sides results in the following: $f(4a+2b)+\frac{1}{2}(f(4a+2c)+f(4b+2a))\ge f(3a+3b)+f(4a+b+c)-f(3a+2b+c)-f(a+2b+3c)$ taking the cyclic sum of this, we get the following: $\sum_{\text{cyc}} (f(4a+2b)+f(2a+4b)-2f(2(a+b+c))) \ge \sum_{\text{cyc}} (f(3b+3c)+f(4a+b+c)-f(3a+2b+c)-f(3a+b+2c))$ Exponentiating and plugging in the definition of $f$ gets use the following: $\prod_{\text{cyc}} (\frac{(P+\frac{1}{4a+2b})(P+\frac{1}{2a+4b})}{(P+\frac{1}{2(a+b+c)})^2}\ge \prod_{\text{cyc}}(\frac{(P+\frac{1}{4a+b+c})(P+\frac{1}{3b+3c})}{(P+\frac{1}{3a+2b+c})(P+\frac{1}{3a+b+2c})})$ Multiplying the top and bottom by 4 inside the product on the left hand side gets the following: $\prod_{\text{cyc}}(\frac{(2P+\frac{1}{2a+b})(2P+\frac{1}{a+2b})}{(2P+\frac{1}{a+b+c})^2})\ge \prod_{\text{cyc}}(\frac{(P+\frac{1}{4a+b+c})(P+\frac{1}{3b+3c})}{(P+\frac{1}{3a+2b+c})(P+\frac{1}{3a+b+2c})})$ Q.E.D. Note: P only needs to be nonnegative for this inequality to hold.