Find the Minimum Value of: $$\frac{18}{a+b} + \frac{12}{ab} + 8a + 5b.$$
where a and b are positive real numbers
I tried evaluating the expression into 1 denominator, and tried to get squares so i can evaluate to 0 but obviously it was the wrong approach.
By AM_GM we obtain: $$\frac{18}{a+b}+\frac{12}{ab}+8a+5b=\left(\frac{18}{a+b}+2(a+b)\right)+\left(\frac{12}{ab}+6a+3b\right)\geq$$ $$\geq2\sqrt{\frac{18}{a+b}\cdot2(a+b)}+3\sqrt[3]{\frac{12}{ab}\cdot6a\cdot3b}=30.$$ The equality occurs for $(a,b)=(1,2)$, which says that $30$ is a minimal value.