We are tasked with showing that if $a, b, c > 0$ and $a + b + c = 3$, then the following inequality holds:

$\frac{a^2 + b^2}{\sqrt{ab}} + \frac{b^2 + c^2}{\sqrt{bc}} + \frac{c^2 + a^2}{\sqrt{ca}} > 6.$

Step 1: Symmetry and Structure of the Inequality

First, notice that this inequality involves the terms $a^2 + b^2$, $b^2 + c^2$, and $c^2 + a^2$, all divided by the square roots of products of the variables, such as $\sqrt{ab}$, $\sqrt{bc}$, and $\sqrt{ca}$. We will aim to exploit the symmetry of this expression and potentially apply well-known inequalities like the AM-GM inequality (Arithmetic Mean-Geometric Mean inequality) to simplify the expression and derive the result.

Step 2: Use of AM-GM Inequality

The AM-GM inequality states that for non-negative real numbers $x$ and $y$,

$\frac{x + y}{2} \geq \sqrt{xy},$

with equality if and only if $x = y$. We will apply this inequality to terms involving $a^2$, $b^2$, and similar terms. Specifically, consider the term $\frac{a^2 + b^2}{\sqrt{ab}}$. Using the AM-GM inequality:

$a^2 + b^2 \geq 2ab,$

and so,

$\frac{a^2 + b^2}{\sqrt{ab}} \geq \frac{2ab}{\sqrt{ab}} = 2\sqrt{ab}.$

Thus, we have the bound:

$\frac{a^2 + b^2}{\sqrt{ab}} \geq 2\sqrt{ab}.$

Similarly, for the other terms, we get:

$\frac{b^2 + c^2}{\sqrt{bc}} \geq 2\sqrt{bc},$ and $\frac{c^2 + a^2}{\sqrt{ca}} \geq 2\sqrt{ca}.$

Step 3: Adding the Inequalities

Summing these inequalities, we obtain:

$\frac{a^2 + b^2}{\sqrt{ab}} + \frac{b^2 + c^2}{\sqrt{bc}} + \frac{c^2 + a^2}{\sqrt{ca}} \geq 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).$

Step 4: Maximizing the Symmetry

Now, we need to show that $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \geq 3$. To do this, we apply the AM-GM inequality again to the terms $\sqrt{ab}$, $\sqrt{bc}$, and $\sqrt{ca}$. By the AM-GM inequality, we have:

[ \frac{\sqrt{ab} + \sqrt{bc} + \sqrt{ca}}{3} \geq \sqrt[3]{\sqrt{ab} \cdot \sqrt{bc} \cdot \sqrt{ca}} = \sqrt[3]{abc}. ]

Since $a + b + c = 3$ and $a, b, c > 0$, we can infer that $abc \leq 1$ by the AM-GM inequality (since $\frac{a + b + c}{3} \geq \sqrt[3]{abc}$).

Thus,

$\sqrt[3]{abc} \leq 1.$

Therefore, we conclude that:

$\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \geq 3.$

Step 5: Final Bound

Using this, we find that:

$\frac{a^2 + b^2}{\sqrt{ab}} + \frac{b^2 + c^2}{\sqrt{bc}} + \frac{c^2 + a^2}{\sqrt{ca}} \geq 2 \times 3 = 6.$

Thus, we have shown that:

$\frac{a^2 + b^2}{\sqrt{ab}} + \frac{b^2 + c^2}{\sqrt{bc}} + \frac{c^2 + a^2}{\sqrt{ca}} > 6,$

as required.

Conclusion

We have demonstrated that the inequality holds, using the AM-GM inequality and the fact that $a + b + c = 3$ implies certain bounds on the terms involving $\sqrt{ab}$, $\sqrt{bc}$, and $\sqrt{ca}$.

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Would you like further details or have any questions about specific steps?

Here are 5 related questions for further exploration:

1. How does the AM-GM inequality apply to inequalities involving sums and products?
2. Can the AM-GM inequality be used for higher powers of variables?
3. What other inequalities could we use to strengthen or simplify this result?
4. How would the inequality change if the condition $a + b + c = 3$ were replaced by $a + b + c = k$?
5. How does symmetry in expressions help in solving inequalities?

Tip: The Arithmetic Mean-Geometric Mean inequality is one of the most powerful tools in inequality proofs, especially when symmetry and products of terms are involved.