Prove The Inequality: A Step-by-Step Guide

$$

Hey guys! Today, we're diving into a fascinating inequality problem that combines elements of real analysis, inequalities, and contest math. Buckle up, because we're about to embark on a mathematical journey to prove that $\prod\limits_{\mathrm{cyc}}(1+\frac{1}{\sqrt{ab}})^a\geq 2^{a+b+c+d}$ for positive real numbers $a, b, c,$ and $d$ where $a+b+c+d \le 4$. This problem is a real gem, and we'll break it down step by step so you can understand the logic and techniques involved. Let's get started!

Understanding the Problem Statement

Before we jump into the proof, let's make sure we fully grasp what the problem is asking. We're given four positive real numbers, $a, b, c,$ and $d$, with a constraint: their sum is less than or equal to 4. The goal is to prove that a specific inequality holds true. This inequality involves a cyclic product (that's what the $\prod\limits_{\mathrm{cyc}}$ notation means) of terms like $(1 + \frac{1}{\sqrt{ab}})^a$ and an exponential term $2^{a+b+c+d}$. To really understand this, let’s expand the cyclic product and see what we’re dealing with:

$$\left(1+\frac{1}{\sqrt{ab}}\right)^a\left(1+\frac{1}{\sqrt{bc}}\right)^b\left(1+\frac{1}{\sqrt{cd}}\right)^c\left(1+\frac{1}{\sqrt{da}}\right)^d \geq 2^{a+b+c+d}$$

So, we need to show that this inequality holds whenever $a, b, c,$ and $d$ are positive and their sum doesn't exceed 4. This is a classic inequality problem, and it often involves clever applications of well-known inequalities like AM-GM (Arithmetic Mean-Geometric Mean) or Jensen's inequality. Our approach will be to leverage these powerful tools to crack this nut.

Key Inequalities and Techniques

To tackle this problem effectively, we'll need some key inequalities in our toolkit. The most important one for this particular problem is the AM-GM inequality. For non-negative numbers $x_1, x_2, ..., x_n$, the AM-GM inequality states that:

$$\frac{x_1 + x_2 + ... + x_n}{n} \geq \sqrt[n]{x_1x_2...x_n}$$

The AM-GM inequality is incredibly versatile and provides a lower bound for the arithmetic mean in terms of the geometric mean. We'll also use the fact that the natural logarithm function, $\ln(x)$, is concave. This property is crucial because it allows us to apply Jensen's inequality. For a concave function $f(x)$ and non-negative weights $\lambda_i$ such that $\sum \lambda_i = 1$, Jensen's inequality states:

$$f(\sum \lambda_i x_i) \geq \sum \lambda_i f(x_i)$$

In simpler terms, for a concave function, the function of the weighted average is greater than or equal to the weighted average of the function values. These two inequalities, AM-GM and Jensen's, will be our main weapons in proving the given inequality. We'll use them strategically to manipulate the terms and arrive at the desired result. The trick is often in how we choose to apply these inequalities – what terms to group together and how to set up the weights for Jensen's inequality.

The Proof: A Step-by-Step Approach

Alright, let's get our hands dirty and dive into the proof. Here's a step-by-step breakdown of how we can demonstrate that the inequality holds true:

Step 1: Applying the Natural Logarithm

Since both sides of the inequality are positive, we can take the natural logarithm (ln) of both sides without changing the direction of the inequality. This is a common trick in inequality problems, as it often simplifies the expressions. Taking the natural logarithm, we get:

$$\ln\left(\prod\limits_{\mathrm{cyc}}\left(1+\frac{1}{\sqrt{ab}}\right)^a\right) \geq \ln\left(2^{a+b+c+d}\right)$$

Using the properties of logarithms, we can rewrite this as:

$$\sum\limits_{\mathrm{cyc}} a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq (a+b+c+d)\ln(2)$$

Step 2: Focusing on a Single Term and Applying AM-GM

Now, let's focus on a single term in the sum, say $a \ln\left(1+\frac{1}{\sqrt{ab}}\right)$. We want to find a lower bound for this term. Here's where the AM-GM inequality comes in handy. We can rewrite the term inside the logarithm as follows:

$$1 + \frac{1}{\sqrt{ab}} = 1 + \frac{1}{2\sqrt{ab}} + \frac{1}{2\sqrt{ab}}$$

Applying AM-GM to these three terms (1, $\frac{1}{2\sqrt{ab}}$, and $\frac{1}{2\sqrt{ab}}$), we get:

$$\frac{1 + \frac{1}{2\sqrt{ab}} + \frac{1}{2\sqrt{ab}}}{3} \geq \sqrt[3]{1 \cdot \frac{1}{2\sqrt{ab}} \cdot \frac{1}{2\sqrt{ab}}} = \sqrt[3]{\frac{1}{4ab}}$$

This simplifies to:

$$1 + \frac{1}{\sqrt{ab}} \geq 3\sqrt[3]{\frac{1}{4ab}}$$

Step 3: Refining the Inequality

However, the above result, while correct, doesn't directly lead us to the desired inequality. We need a slightly different approach. Let’s go back to the original term $a \ln\left(1+\frac{1}{\sqrt{ab}}\right)$ and use a simpler bound. Since $1 + x \geq 2\sqrt{x}$ for $x = \frac{1}{\sqrt{ab}}$, we have:

$$1 + \frac{1}{\sqrt{ab}} \geq 2\sqrt{\frac{1}{\sqrt{ab}}} = 2(ab)^{-\frac{1}{4}}$$

Therefore:

$$a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq a \ln\left(2(ab)^{-\frac{1}{4}}\right) = a \left(\ln(2) - \frac{1}{4}\ln(ab)\right) = a\ln(2) - \frac{a}{4}(\ln(a)+\ln(b))$$

Step 4: Summing Cyclically and Simplifying

Now, let's sum this inequality cyclically over the four terms:

$$\sum\limits_{\mathrm{cyc}} a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq \sum\limits_{\mathrm{cyc}} \left(a\ln(2) - \frac{a}{4}(\ln(a)+\ln(b))\right)$$

The left-hand side is exactly what we want. Let's simplify the right-hand side:

$$\sum\limits_{\mathrm{cyc}} a\ln(2) = (a+b+c+d)\ln(2)$$

For the second part of the sum, we have:

$$\sum\limits_{\mathrm{cyc}} \frac{a}{4}(\ln(a)+\ln(b)) = \frac{1}{4}(a\ln(a) + a\ln(b) + b\ln(b) + b\ln(c) + c\ln(c) + c\ln(d) + d\ln(d) + d\ln(a))$$

This can be rearranged as:

$$\frac{1}{4}((a+d)\ln(a) + (a+b)\ln(b) + (b+c)\ln(c) + (c+d)\ln(d))$$

So, our inequality now looks like:

$$\sum\limits_{\mathrm{cyc}} a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq (a+b+c+d)\ln(2) - \frac{1}{4}((a+d)\ln(a) + (a+b)\ln(b) + (b+c)\ln(c) + (c+d)\ln(d))$$

Step 5: Applying Jensen's Inequality (Here's where it gets interesting!)

To proceed further, we need to show that the term we subtracted is non-positive. This is where Jensen's inequality comes into play. Let's consider the function $f(x) = x\ln(x)$, which is convex for $x > 0$. We want to show that:

$$(a+d)\ln(a) + (a+b)\ln(b) + (b+c)\ln(c) + (c+d)\ln(d) \leq 4\ln(2)$$

This is where we might hit a roadblock! Applying Jensen’s directly isn’t straightforward here, and this is where the problem’s difficulty truly lies. The inequality isn’t as clean as it initially seemed.

Step 6: A More Refined Approach (The Key Insight)

Instead of forcing Jensen's inequality, we need a more insightful bound. Let's go back to Step 3 and refine our approach. We had:

$$a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq a\ln(2) - \frac{a}{4}(\ln(a)+\ln(b))$$

The issue is that the $-\frac{a}{4}(\ln(a)+\ln(b))$ term is tricky to handle directly. Let’s try a different lower bound for the logarithm. Using the inequality $\ln(1+x) \geq \frac{x}{1+x}$ for $x > 0$, we have:

$$\ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq \frac{\frac{1}{\sqrt{ab}}}{1 + \frac{1}{\sqrt{ab}}} = \frac{1}{\sqrt{ab} + 1}$$

So:

$$a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq \frac{a}{1+\sqrt{ab}}$$

Step 7: Applying AM-GM Again (A Clever Twist)

Now, we apply AM-GM to the denominator: $1 + \sqrt{ab} \leq 1 + \frac{a+b}{2}$. Thus:

$$\frac{a}{1+\sqrt{ab}} \geq \frac{a}{1+\frac{a+b}{2}} = \frac{2a}{2+a+b}$$

Step 8: Summing Cyclically (The Magic Happens)

Summing cyclically, we get:

$$\sum\limits_{\mathrm{cyc}} a \ln\left(1+\frac{1}{\sqrt{ab}}\right) \geq \sum\limits_{\mathrm{cyc}} \frac{2a}{2+a+b}$$

Step 9: A Final Transformation (The Home Stretch)

Now, this is where the magic happens! We need to show that:

$$\sum\limits_{\mathrm{cyc}} \frac{2a}{2+a+b} \geq (a+b+c+d)\ln(2)$$

This inequality is still not trivial, but it's more manageable. This is a classic inequality pattern which reminds us of the Shapiro inequality (though this isn’t directly Shapiro). To solve this, we require further manipulations or a more refined inequality. However, the core techniques—using AM-GM, strategically bounding logarithms, and cyclic summation—are the key ingredients. While a complete, clean solution requires more advanced techniques, this breakdown illustrates the typical problem-solving process in contest math: try different approaches, refine your bounds, and leverage key inequalities.

Conclusion: The Art of Inequality Problem Solving

Proving inequalities like this one is a challenging but rewarding endeavor. The key takeaways are the strategic application of inequalities like AM-GM and Jensen's, the clever manipulation of terms, and the importance of trying different approaches when you hit a roadblock. This particular problem highlights that sometimes the most obvious path isn't the correct one, and you need to dig deeper and find a more refined approach. While we haven't presented a complete, polished solution here, the steps we've outlined demonstrate the thought process and techniques involved in tackling such problems. Keep practicing, keep exploring, and you'll become a master of inequalities in no time!