<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] we get $$\frac{a+c}{b+c}-\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}=-{\frac { \left( 3\,c+d \right)  \left( a-b \right)  \left( abc+abd+ca
d+cbd \right)  \left( c+a \right) }{ \left( 3\,{a}^{2}{b}^{2}c+3\,{a}^
{2}{b}^{2}d+3\,{a}^{2}b{c}^{2}+4\,{a}^{2}bcd+2\,{a}^{2}b{d}^{2}+3\,{a}
^{2}{c}^{2}d+2\,{a}^{2}c{d}^{2}+6\,a{b}^{2}{c}^{2}+8\,a{b}^{2}cd+3\,a{
b}^{2}{d}^{2}+8\,ab{c}^{2}d+4\,abc{d}^{2}+3\,a{c}^{2}{d}^{2}+6\,{b}^{2
}{c}^{2}d+3\,{b}^{2}c{d}^{2}+3\,b{c}^{2}{d}^{2} \right)  \left( c+b
 \right) }}
$$
and this is positive since we have $$a\le b$$
</s>