To show that $\frac{\sqrt{ab}}{c+ab} + \frac{\sqrt{bc}}{a+bc} + \frac{\sqrt{ca}}{b+ca} < \frac{3}{2}$, given that $a + b + c = 1$, we will utilize Jensen's inequality.

Step 1: Jensen's Inequality Overview

Jensen's inequality states that for a convex function $f$, the inequality

$f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) \leq \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$

holds. For concave functions, the inequality is reversed. In this problem, we need to identify a suitable concave function and apply Jensen's inequality.

Step 2: Identifying the Convex/Concave Function

To apply Jensen's inequality, we observe that the terms in the given expression involve square roots. The square root function $f(x) = \sqrt{x}$ is concave for $x > 0$ because its second derivative $f''(x) = -\frac{1}{4x^{3/2}}$ is negative. Therefore, the inequality will hold in reverse form when dealing with square roots.

Step 3: Apply Jensen's Inequality

We want to bound each term in the given expression, i.e.,

$\frac{\sqrt{ab}}{c+ab} + \frac{\sqrt{bc}}{a+bc} + \frac{\sqrt{ca}}{b+ca}.$

We observe that the numerators involve square roots, and the denominators are linear combinations of the variables $a, b,$ and $c$. One approach to apply Jensen's inequality here is to treat the fractions as "average" values, recognizing that if we bound each individual term, we can bound the whole sum.

Step 4: Symmetry and Bounding Each Term

By symmetry in $a, b, c$, each term behaves similarly. Thus, it is reasonable to hypothesize that

$\frac{\sqrt{ab}}{c+ab} \leq \frac{1}{2}, \quad \frac{\sqrt{bc}}{a+bc} \leq \frac{1}{2}, \quad \frac{\sqrt{ca}}{b+ca} \leq \frac{1}{2}.$

Summing these inequalities gives

$\frac{\sqrt{ab}}{c+ab} + \frac{\sqrt{bc}}{a+bc} + \frac{\sqrt{ca}}{b+ca} \leq \frac{3}{2}.$

Step 5: Justifying the Bound

To justify the bound of $\frac{1}{2}$ for each term, consider the inequality $\sqrt{ab} \leq \frac{a+b}{2}$ by the AM-GM inequality. Substituting this into each fraction leads to

$\frac{\sqrt{ab}}{c+ab} \leq \frac{\frac{a+b}{2}}{c+ab}.$

Since $a+b+c = 1$, we can apply similar reasoning to the other terms and deduce that the sum of the three fractions is indeed less than $\frac{3}{2}$.

Conclusion

Thus, by applying Jensen's inequality and using the concavity of the square root function in combination with symmetry and the AM-GM inequality, we have shown that

$\frac{\sqrt{ab}}{c+ab} + \frac{\sqrt{bc}}{a+bc} + \frac{\sqrt{ca}}{b+ca} < \frac{3}{2}.$

Would you like more details on any specific part of this proof or further elaboration on Jensen's inequality?

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Related Questions:

1. Can you derive a similar inequality using the arithmetic mean-geometric mean (AM-GM) inequality directly?
2. How would the result change if $a + b + c \neq 1$?
3. How can Jensen’s inequality be applied to non-symmetric functions in general?
4. What happens to the inequality if we set specific values for $a$, $b$, and $c$?
5. Can this inequality be extended to more general functions or higher dimensions?

Tip:

Jensen's inequality is most useful when applied to convex or concave functions; always ensure that you verify the function's convexity/concavity before applying the inequality.