Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$

Question

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

Answer 1

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

Answer 2

Hint: $(a+b-c)^2 \ge 0 \implies (a^2+b^2+c^2) \ge 2(bc + ca - ab)$

Can you relate both sides to the question at hand?