Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$ $a;b;c>0$ such that $a^2+b^2+c^2=\frac{5}{3}$. Prove :
$\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$

$$
c(a+b)\leq \frac{c^2+(a+b)^2}{2}=\frac{a^2+b^2+c^2}{2}+ab=\frac 56+ab
$$
which is the rearrangement of your inequality and even the stronger one. 
$$
c(a+b)\leq \frac 56+ab \implies \\
\frac 1a+\frac 1b\leq \frac 5{6abc}+\frac 1c\implies
\frac 1a+\frac 1b-\frac 1c\leq \frac 5{6abc}
$$
