Proving Inequalities for Positive Real Numbers a, b, and c

In this article, we explore a proof for the inequality involving positive real numbers a, b, and c, specifically that the following holds unless a b c:

(frac{a}{bc} frac{b}{ca} frac{c}{ab} geq frac{3}{2})

We will use several well-known inequalities, including Titus lemma, Cauchy-Schwarz inequality, and the A.M.-G.M. inequality.

Step-by-Step Proof

Step 1: Applying Titus Lemma

To begin, we apply the Titus lemma, which states that for positive real numbers (a, b, c > 0):

(frac{a^2}{abc} frac{b^2}{bca} frac{c^2}{cab} geq frac{abc^2}{ab ac bc})

Substituting the given fractions, we have:

(frac{a}{bc} frac{b}{ca} frac{c}{ab} geq frac{abc}{ab ac bc})

Step 2: Analyzing the Expression

We need to show that:

(frac{abc}{ab ac bc} geq frac{3}{2})

This rearranges to:

(2abc geq 3(ab ac bc))

Step 3: Using the Cauchy-Schwarz Inequality

Using the Cauchy-Schwarz inequality, we know:

(abc^2 geq 3ab ac bc)

Equality holds if and only if (a b c).

Therefore, the inequality (abc^2 geq 3ab ac bc) holds true unless (a b c).

Step 4: Conclusion

Combining these results, we conclude that:

(frac{a}{bc} frac{b}{ca} frac{c}{ab} geq frac{3}{2})

unless (a b c).

Alternative Solution Using Isolated Fudging

For a more unique approach, let's consider an alternative proof inspired by isolated fudging. The basic idea here is to find a suitable positive real number (r) such that:

(frac{a}{bc} geq frac{3}{2} cdot frac{a^r}{a^r b^r c^r})

Similarly, we need:

(frac{b}{ca} geq frac{3}{2} cdot frac{b^r}{a^r b^r c^r})

and

(frac{c}{ab} geq frac{3}{2} cdot frac{c^r}{a^r b^r c^r})

Adding these three inequalities, we get the desired result.

Finding Suitable r

We start by manipulating the first inequality:

(frac{a}{bc} geq frac{3}{2} cdot frac{a^r}{a^r b^r c^r})

This is equivalent to:

(2a^{r-1} b^{r-1} c^{r-1} geq 3)

To ensure this holds true under the condition that (a b c), we need:

(2a^{2r-2} c geq 3a^{2r-2})

This simplifies to:

(2a^{2r-2} c geq 3a^{2r-2})

Hence, we need to find (r) such that:

(2a^{2r-2} c geq 3a^{2r-2})

By inspection, (r frac{3}{2}) satisfies both conditions.

Substituting (r frac{3}{2}), we get:

(2a^{frac{3}{2} - 1} b^{frac{3}{2} - 1} c^{frac{3}{2} - 1} 2a^{frac{1}{2}} b^{frac{1}{2}} c^{frac{1}{2}} geq 3a^{frac{3}{2} - 1} b^{frac{1}{2}} c^{frac{3}{2} - 1})

Thus, the inequalities hold, and summing them, we get:

(frac{a}{bc} frac{b}{ca} frac{c}{ab} geq frac{3}{2})

Therefore, the proof is complete.