prove that if a+b+c>abc then prove a^2+b^2+c^2>√3 abc

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 >

Prove that if a + b + c > abc then prove a^2 + b^2 + c^2 > √3 abc

Explore a detailed proof for the inequality a^2 + b^2 + c^2 > √3 abc, given that a + b + c > abc. This problem involves concepts of inequalities, including the AM-GM and Cauchy-Schwarz inequalities, making it suitable for advanced high school students in Grades 11-12.

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