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.

I found a proof by Titu's Lemma$:$
Let $$\text{LHS} -\text{RHS} \equiv \frac{g(a,b,c)}{4abc(a+b+c)^2}$$
But we have
\begin{align*} g(a,\,b,\,c) &=\frac{1}{2} \sum\limits_{cyc} c^2(a+b-5c)^2 (a-b)^2-9(a-b)^2(b-c)^2(c-a)^2 \\&\geq \frac{1}{2(a^2+b^2+c^2)} \Big[\sum\limits_{cyc} c^2(a+b-5c)(a-b)\Big]^2-9(a-b)^2(b-c)^2(c-a)^2\\&={\frac { \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( c-a
 \right) ^{2} \left( 7\,{a}^{2}+50\,ab+50\,ac+7\,{b}^{2}+50\,bc+7\,{c}
^{2} \right) }{2\,(a^2+b^2+c^2)}} \geq 0\end{align*}
So we are done.
