<s>[INST] Inequality with positive reals For each $x,y,z>0$, define 
$$
f(x,y,z)=\frac{3x^2(y+z)+2xyz}{(x+y)(y+z)(z+x)}.
$$
Fix also $a,b,c,d>0$ with $a\le b$. How can we prove "reasonably" the following inequality?
$$
\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)} \le \frac{a+c}{b+c}.
$$
 [/INST] The expression
$$\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}
 \le \frac{a+c}{b+c}$$ is equivalent to
$$(b+c)(f(a,b,d)+f(c,b,d))\le(a+c)(f(b,a,d)+f(c,a,d))$$ The $LHS$ is equal to
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2bcd}{(b+d)(c+d)}$$
The $RHS$ is equal to
$$\frac{3b^2(a+c)}{(a+b)(b+d)}+\frac{2bda^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2acd}{(a+d)(c+d)}$$ so we have to prove that 
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2}{(a+b)(a+d)(b+d)}+\frac{2bcd}{(b+d)(c+d)}$$ is less or equal than
$$\frac{3b^2(a+c)}{(a+b)(b+d)}+\frac{2bda^2}{(a+b)(a+d)(b+d)}+\frac{2acd}{(a+d)(c+d)}$$ Multiplying by $(a+b)(a+d)(b+d)(c+d)$ one has respectively
$$(b+d)(c+d)(b+c)(3a^2)+(c+d)(2adb^2)+(a+b)(a+d)(2bcd)$$ and
$$(a+d)(a+c)(c+d)(3b^2)+(c+d)(2bda^2)+(a+b)(b+d)(2acd)$$
What precedes becomes after simplification $$a^2(b+c)(b+d)\le b^2(a+c)(a+d)$$ where $a,b,c,d\gt 0$ and $a\le b$.
Finally we have the evident inequality
$$\color{red}{0\le ab(b-a)(c+d)+(b^2-a^2)cd}$$
</s>