Prove The Inequality: A Step-by-Step Guide

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Hey guys! Today, we're diving deep into a fascinating inequality problem that involves square roots, fractions, and a touch of elegance. The problem states: Given non-negative real numbers $a, b, c$ such that $a + b + c = 1$, prove that

$$\sqrt{\frac{5a^2+4}{3b+1}}+\sqrt{\frac{5b^2+4}{3c+1}}+\sqrt{\frac{5c^2+4}{3a+1}}\leq6.$$

Equality holds if and only if $a = b = 1$ and $c = 0$. Sounds intriguing, right? Let's break it down step by step and explore the beautiful world of inequalities together!

Understanding the Problem

Before we jump into solving this inequality, let's take a moment to really understand what it's asking. We're dealing with three variables, $a$, $b$, and $c$, which are non-negative real numbers. This means they can be zero or any positive number. The constraint $a + b + c = 1$ tells us that these numbers must add up to 1, which is a crucial piece of information. This constraint actually confines our variables to a specific region in 3D space, which can be helpful in visualizing the problem.

The inequality itself looks a bit intimidating at first glance. We have square roots, fractions with quadratic expressions in the numerator ($5a^2 + 4$, $5b^2 + 4$, $5c^2 + 4$), and linear expressions in the denominator ($3b + 1$, $3c + 1$, $3a + 1$). The goal is to show that the sum of these three square root terms is always less than or equal to 6. The condition for equality, $a = b = 1$ and $c = 0$, gives us a clue about where the maximum value of the expression might occur. However, note a typo in the original problem: it should likely state that equality holds when two of the variables are $\frac{1}{2}$ and the other is 0, which we'll discuss further as we solve it.

To tackle this, we'll need to employ some clever techniques and inequalities that are our trusty tools in the world of mathematical problem-solving. Think Cauchy-Schwarz, AM-GM, or maybe even some clever algebraic manipulation. Let’s roll up our sleeves and get started!

Strategizing a Solution: Key Inequalities and Techniques

Okay, so we've got this beast of an inequality staring us down. The next step is to formulate a plan of attack. Inequalities like these often require a strategic combination of techniques. Here's what we're thinking:

1. Cauchy-Schwarz Inequality: This is a classic workhorse for inequality problems, especially those involving sums of products. It might help us to separate the square roots and create more manageable terms.
2. AM-GM Inequality: The Arithmetic Mean-Geometric Mean (AM-GM) inequality is another powerful tool that relates the average of numbers to their product. It could be useful for bounding the terms inside the square roots or the denominators.
3. Algebraic Manipulation: Sometimes, simply rearranging the terms or adding/subtracting clever expressions can reveal hidden structures and lead to a solution.
4. Symmetric Sums: Recognizing the symmetry in the problem (the variables $a$, $b$, and $c$ are interchangeable) can help us simplify the expressions and focus on a single variable at a time.

Given the structure of the inequality, the Cauchy-Schwarz inequality seems like a promising starting point. Remember, the Cauchy-Schwarz inequality states that for any real numbers $x_i$ and $y_i$:

$$(x_1^2 + x_2^2 + ... + x_n^2)(y_1^2 + y_2^2 + ... + y_n^2) \geq (x_1y_1 + x_2y_2 + ... + x_ny_n)^2$$

We need to figure out how to apply this to our problem. The square roots suggest that we might want to consider terms like $\sqrt{\frac{5a^2+4}{3b+1}}$ as our $x_i$ terms. But what should our $y_i$ terms be? This is where the creativity and problem-solving intuition come into play! We'll explore this in the next section.

Applying Cauchy-Schwarz: A First Step

Let's put our Cauchy-Schwarz strategy into action. We have the terms $\sqrt{\frac{5a^2+4}{3b+1}}$, $\sqrt{\frac{5b^2+4}{3c+1}}$, and $\sqrt{\frac{5c^2+4}{3a+1}}$. If we consider these as our $x_i$ terms, we need to find suitable $y_i$ terms that will help us simplify the expression. A common trick when dealing with fractions inside square roots is to choose $y_i$ terms that involve the denominator of the fraction.

So, let's try setting:

• $x_1 = \sqrt{\frac{5a^2+4}{3b+1}}$, $x_2 = \sqrt{\frac{5b^2+4}{3c+1}}$, $x_3 = \sqrt{\frac{5c^2+4}{3a+1}}$
• $y_1 = \sqrt{3b+1}$, $y_2 = \sqrt{3c+1}$, $y_3 = \sqrt{3a+1}$

Now, let's apply the Cauchy-Schwarz inequality:

(\sqrt{\frac{5a^2+4}{3b+1}}^2 + \sqrt{\frac{5b^2+4}{3c+1}}^2 + \sqrt{\frac{5c^2+4}{3a+1}}^2)(\sqrt{3b+1}^2 + \sqrt{3c+1}^2 + \sqrt{3a+1}^2) \geq$ $(\sqrt{\frac{5a^2+4}{3b+1}} \cdot \sqrt{3b+1} + \sqrt{\frac{5b^2+4}{3c+1}} \cdot \sqrt{3c+1} + \sqrt{\frac{5c^2+4}{3a+1}} \cdot \sqrt{3a+1})^2

Simplifying this, we get:

$$(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1})(3b+1 + 3c+1 + 3a+1) \geq (\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2$$

Since $a + b + c = 1$, we have $3a + 3b + 3c = 3$, so the second term on the left-hand side simplifies to $3 + 3 = 6$. Thus,

$$(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1}) \cdot 6 \geq (\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2$$

This is a good step forward! We've managed to get rid of the square roots in the denominators. However, we're still left with square roots on the right-hand side and fractions on the left-hand side. We need to find a way to bound these terms further. Let's think about how we can simplify the expression $\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4}$.

Bounding $\sqrt{5a^2+4}$: A Crucial Inequality

The expression $\sqrt{5a^2 + 4}$ is a key piece of the puzzle. To make progress, we need to find a good upper bound for this term. Notice that the expression is a function of $a^2$. A linear upper bound would be ideal, as it would simplify the sum $\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4}$.

Let's consider a linear function of the form $Aa + B$, where $A$ and $B$ are constants. We want to find values for $A$ and $B$ such that

$$\sqrt{5a^2 + 4} \leq Aa + B$$

for all $a$ in the interval $[0, 1]$ (since $a, b, c$ are non-negative and $a + b + c = 1$). To find appropriate values for $A$ and $B$, we can consider the endpoints of the interval and a strategic midpoint. Let's look at $a = 0$, $a = 1$, and $a = \frac{1}{2}$:

• For $a = 0$, we have $\sqrt{5(0)^2 + 4} = 2$. So, we need $B \geq 2$.
• For $a = 1$, we have $\sqrt{5(1)^2 + 4} = \sqrt{9} = 3$. So, we need $A + B \geq 3$.
• For $a = \frac{1}{2}$, we have $\sqrt{5(\frac{1}{2})^2 + 4} = \sqrt{\frac{5}{4} + 4} = \sqrt{\frac{21}{4}} = \frac{\sqrt{21}}{2} \approx 2.29$. So, we need $\frac{1}{2}A + B \geq \frac{\sqrt{21}}{2}$.

Let's try $A = 1$ and $B = 2$. Then, our inequality becomes $\sqrt{5a^2 + 4} \leq a + 2$. We can verify if this inequality holds by squaring both sides:

$$5a^2 + 4 \leq (a + 2)^2 = a^2 + 4a + 4$$

$$4a^2 - 4a \leq 0$$

$$4a(a - 1) \leq 0$$

Since $0 \leq a \leq 1$, the inequality $4a(a - 1) \leq 0$ is indeed true! Therefore, we have found a crucial upper bound:

$$\sqrt{5a^2 + 4} \leq a + 2$$

This is a fantastic result! We can now use this inequality to bound the sum $\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4}$.

Summing the Inequalities and Reaching the Final Bound

Now that we have the inequality $\sqrt{5a^2 + 4} \leq a + 2$, we can apply the same logic to $b$ and $c$ to get:

$$\sqrt{5b^2 + 4} \leq b + 2$$

$$\sqrt{5c^2 + 4} \leq c + 2$$

Adding these three inequalities together, we get:

$$\sqrt{5a^2 + 4} + \sqrt{5b^2 + 4} + \sqrt{5c^2 + 4} \leq (a + 2) + (b + 2) + (c + 2)$$

Since $a + b + c = 1$, this simplifies to:

$$\sqrt{5a^2 + 4} + \sqrt{5b^2 + 4} + \sqrt{5c^2 + 4} \leq 1 + 6 = 7$$

Now, let's go back to the Cauchy-Schwarz inequality we derived earlier:

$$(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1}) \cdot 6 \geq (\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2$$

We know that $\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4} \leq 7$, so:

$$(\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2 \leq 7^2 = 49$$

Thus,

$$(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1}) \cdot 6 \geq 49$$

$$\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1} \geq \frac{49}{6}$$

This inequality, while interesting, doesn't directly lead us to the desired result. We need to find another way to connect this back to the original inequality.

Let's revisit our original goal: to prove that

$$\sqrt{\frac{5a^2+4}{3b+1}}+\sqrt{\frac{5b^2+4}{3c+1}}+\sqrt{\frac{5c^2+4}{3a+1}}\leq6$$

We can try applying the Cauchy-Schwarz inequality in a different way. Consider the vectors:

$$u = \left(\sqrt{\frac{5a^2+4}{3b+1}}, \sqrt{\frac{5b^2+4}{3c+1}}, \sqrt{\frac{5c^2+4}{3a+1}}\right)$$

$$v = (\sqrt{3b+1}, \sqrt{3c+1}, \sqrt{3a+1})$$

By Cauchy-Schwarz,

$$(u \cdot u)(v \cdot v) \geq (u \cdot v)^2$$

$$(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1})(3b+1 + 3c+1 + 3a+1) \geq (\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2$$

We already know $3b+1 + 3c+1 + 3a+1 = 3(a+b+c) + 3 = 6$. Let $X = \sqrt{\frac{5a^2+4}{3b+1}}+\sqrt{\frac{5b^2+4}{3c+1}}+\sqrt{\frac{5c^2+4}{3a+1}}$. We want to show $X \leq 6$.

Applying Cauchy-Schwarz to the original expression directly, let:

$$x_i = \sqrt{\frac{5a^2 + 4}{3b + 1}}, \sqrt{\frac{5b^2 + 4}{3c + 1}}, \sqrt{\frac{5c^2 + 4}{3a + 1}}$$

$$y_i = 1, 1, 1$$

Then,

$$(x_1^2 + x_2^2 + x_3^2)(y_1^2 + y_2^2 + y_3^2) \geq (x_1y_1 + x_2y_2 + x_3y_3)^2$$

$$(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1})(1^2 + 1^2 + 1^2) \geq (\sqrt{\frac{5a^2+4}{3b+1}} + \sqrt{\frac{5b^2+4}{3c+1}} + \sqrt{\frac{5c^2+4}{3a+1}})^2$$

$$3(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1}) \geq X^2$$

We still need to bound the term $\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1}$.

Applying Titu's Lemma (Engel's Form)

Titu's Lemma, also known as Engel's form of Cauchy-Schwarz, is very useful for inequalities involving fractions. It states that for positive real numbers $x_i$ and $y_i$:

$$\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + ... + \frac{x_n^2}{y_n} \geq \frac{(x_1 + x_2 + ... + x_n)^2}{y_1 + y_2 + ... + y_n}$$

Applying Titu's Lemma to our expression, we get:

$$\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1} \geq \frac{(\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2}{3b+1 + 3c+1 + 3a+1}$$

We know $3b+1 + 3c+1 + 3a+1 = 6$, and we also know $\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4} \leq a+2 + b+2 + c+2 = 1 + 6 = 7$. Therefore,

$$\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1} \geq \frac{(\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4})^2}{6}$$

Using our earlier bound, $\sqrt{5a^2+4} + \sqrt{5b^2+4} + \sqrt{5c^2+4} \leq 7$, we have:

$$\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1} \geq \frac{7^2}{6} = \frac{49}{6}$$

Going back to the inequality $3(\frac{5a^2+4}{3b+1} + \frac{5b^2+4}{3c+1} + \frac{5c^2+4}{3a+1}) \geq X^2$ we get:

$$3 \cdot \frac{49}{6} \geq X^2$$

$$\frac{49}{2} \geq X^2$$

$$X \leq \sqrt{\frac{49}{2}} = \frac{7}{\sqrt{2}} \approx 4.95$$

This bound is better, but it's still not 6. We need to refine our approach further.

Refining the Bounding: A Closer Look at the Denominators

It seems our initial approach, while utilizing powerful inequalities, isn't giving us the tightest bound. Let's go back to the original inequality and think about the denominators more carefully. We have $3b + 1$, $3c + 1$, and $3a + 1$ in the denominators. Since $a, b, c$ are non-negative and $a + b + c = 1$, these denominators are all between 1 and 4.

The original problem stated that equality holds when $a=b=1$ and $c=0$, but this clearly violates the condition $a+b+c=1$. The correct equality condition occurs when two of the variables are equal to $\frac{1}{2}$ and the other is 0. Without loss of generality, let's consider the case $a = \frac{1}{2}$, $b = \frac{1}{2}$, and $c = 0$. In this case, the left-hand side of the inequality becomes:

$$\sqrt{\frac{5(\frac{1}{2})^2+4}{3(\frac{1}{2})+1}} + \sqrt{\frac{5(\frac{1}{2})^2+4}{3(0)+1}} + \sqrt{\frac{5(0)^2+4}{3(\frac{1}{2})+1}} = \sqrt{\frac{\frac{5}{4}+4}{\frac{3}{2}+1}} + \sqrt{\frac{\frac{5}{4}+4}{1}} + \sqrt{\frac{4}{\frac{3}{2}+1}}$$

$$= \sqrt{\frac{\frac{21}{4}}{\frac{5}{2}}} + \sqrt{\frac{21}{4}} + \sqrt{\frac{4}{\frac{5}{2}}} = \sqrt{\frac{21}{4} \cdot \frac{2}{5}} + \frac{\sqrt{21}}{2} + \sqrt{\frac{8}{5}} = \sqrt{\frac{21}{10}} + \frac{\sqrt{21}}{2} + \sqrt{\frac{8}{5}}$$

$$\approx 1.45 + 2.29 + 1.26 \approx 5$$

This suggests our bound of 6 might be achievable. To proceed, we might need to consider another approach or a more refined bounding technique.

Let's think about Holder's inequality, which is a generalization of Cauchy-Schwarz. However, before we jump into more complex inequalities, let's try a simpler approach. We've established that $\sqrt{5a^2 + 4} \leq a + 2$. Can we use this to our advantage in a different way?

Let's go back to the original inequality and multiply each square root term by 1 in a clever way:

$$\sqrt{\frac{5a^2+4}{3b+1}} = \sqrt{\frac{5a^2+4}{3b+1}} \cdot \frac{\sqrt{3b+1}}{\sqrt{3b+1}} = \frac{\sqrt{(5a^2+4)(3b+1)}}{3b+1}$$

Now, we can apply the inequality $\sqrt{5a^2 + 4} \leq a + 2$ inside the square root. However, this might complicate things further. Let's try a different approach.

Conclusion (Incomplete Solution and Further Steps)

We've made significant progress in analyzing this inequality. We've applied Cauchy-Schwarz in various ways, derived a crucial upper bound for $\sqrt{5a^2 + 4}$, and explored the use of Titu's Lemma. However, we haven't yet arrived at the final proof that

$$\sqrt{\frac{5a^2+4}{3b+1}}+\sqrt{\frac{5b^2+4}{3c+1}}+\sqrt{\frac{5c^2+4}{3a+1}}\leq6$$

Further steps might involve:

1. Exploring Holder's Inequality: This could provide a more powerful tool for dealing with the sum of square roots.
2. Refining the Bound for $\sqrt{5a^2 + 4}$: Perhaps a tighter linear bound can be found.
3. Analyzing the Equality Condition More Carefully: Understanding when equality holds can give us valuable clues about the behavior of the inequality.
4. Trying a Different Approach Entirely: Sometimes, a fresh perspective is needed to crack a tough problem.

This problem is a great example of how inequality problems can require a combination of techniques and careful analysis. We've laid a strong foundation for solving it, and with a few more clever steps, we can likely reach the finish line. Keep exploring, keep experimenting, and most importantly, keep enjoying the journey of mathematical discovery! Remember that math is not about getting the answer right away, it's about the process. Good luck, and happy problem-solving!