<s>[INST] Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$ For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$
Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left( 
a+b-2\,c \right) ^{2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left( 
119\,a+119\,b+30\,c \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2}
 \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right)  \left( a+b \right)  \left( 16\,{a}^{2}+a
b+16\,{b}^{2} \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right)  \left( a+b \right) ^{
2} \Big] \geqq 0$
which is clearly true for $c=\max\{a,b,c\}$
I wish to see another proof without $uvw$! Thanks for a real lot!
You can see also here.
 [/INST] BW helps!
Let $a\leq b\leq c$, $b=a+u$ and $c=a+u+v.$
Thus, we need to prove that:
$$9(u^2+uv+v^2)a^4+3(14u^3+21u^2v+3uv^2-2v^3)a^2+$$
$$+(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)a^2+$$
$$+u(56u^4+140u^3v+114u^2v^2+31uv^3+v^4)a+$$
$$+4u^2(2u^2+3uv+v^2)^2\geq0,$$
for which it's enough to prove that:
$$4(u^2+uv+v^2)(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)\geq(14u^3+21u^2v+3uv^2-2v^3)^2$$ or
$$292u^6+876u^5v-85u^4v^2+610u^3v^3+391u^2v^4+8uv^5+20v^6\geq0,$$ which is obvious. 
</s>