To prove that if $a + b + c > abc$, then $a^2 + b^2 + c^2 > \sqrt{3} \cdot abc$, let's work through the conditions carefully. We are assuming that $a$, $b$, and $c$ are positive real numbers, as otherwise the inequality involving $abc$ might not be meaningful.

Proof

1. Step 1: Analyze the Given Condition We start with the assumption: $a + b + c > abc$ We need to show that under this condition, $a^2 + b^2 + c^2 > \sqrt{3} \cdot abc.$

2. Step 2: Apply the Arithmetic Mean-Geometric Mean (AM-GM) Inequality By the AM-GM inequality, for any non-negative real numbers $a$, $b$, and $c$, $a + b + c \geq \sqrt{3 \cdot (ab + bc + ca)}.$ Since $a + b + c > abc$, this inequality suggests that $abc$ is constrained by $a$, $b$, and $c$.

3. Step 3: Use Cauchy-Schwarz Inequality on the Left-Hand Side Another useful inequality is Cauchy-Schwarz, which tells us that: $(a^2 + b^2 + c^2)(1^2 + 1^2 + 1^2) \geq (a + b + c)^2.$ Simplifying $1^2 + 1^2 + 1^2 = 3$, we get: $3(a^2 + b^2 + c^2) \geq (a + b + c)^2.$ Dividing both sides by 3 gives: $a^2 + b^2 + c^2 \geq \frac{(a + b + c)^2}{3}.$

4. Step 4: Relate $a^2 + b^2 + c^2$ to $abc$ Since $a + b + c > abc$, we see that: $\frac{(a + b + c)^2}{3} > \frac{(abc)^2}{3}.$ This implies that: [ a^2 + b^2 + c^2 >