Prove The Inequality: Σ√(x²+yz+2) ≥ 6 | Step-by-Step

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Hey guys! Today, we're diving into a fascinating inequality problem that looks deceptively simple but requires a clever approach to crack. We're going to explore the inequality $\sum\limits_{\mathrm{cyc}}\sqrt{x^2+yz+2}\ge 6$ where $x, y, z \ge 0$ and $x+y+z=3$. This problem falls under the realm of algebra and precalculus, specifically focusing on inequalities. So, buckle up and let's get started!

Problem Statement: Laying the Foundation

The problem statement is crystal clear: Given non-negative real numbers $x, y, z$ such that their sum is 3, we need to prove that the sum of the square roots of the expressions $x^2+yz+2$, $y^2+zx+2$, and $z^2+xy+2$ is greater than or equal to 6. Mathematically, this is expressed as:

$$\sum\limits_{\mathrm{cyc}}\sqrt{x^2+yz+2} = \sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$

This inequality hints at the use of several techniques, including but not limited to, Cauchy-Schwarz Inequality, Minkowski's Inequality, or even clever algebraic manipulations. The presence of the square roots often nudges us towards inequalities that deal well with square roots and sums.

Initial Thoughts and Strategies: Charting Our Course

When tackling such problems, it's crucial to have a game plan. Here are some initial thoughts and strategies we might consider:

1. Simplification: Can we simplify the expression inside the square root? Perhaps by using the given condition $x+y+z=3$, we can rewrite $x^2+yz+2$ in a more manageable form. Remember, $2$ can be expressed as $2 \cdot 1$, and we might be able to relate that to $(x+y+z)$.
2. Inequality Selection: Which inequality should we use? The structure of the problem suggests Cauchy-Schwarz or Minkowski. These inequalities are powerful tools for dealing with sums of square roots. We'll need to carefully examine which one fits best.
3. Homogenization/Normalization: Sometimes, it's beneficial to homogenize the inequality (make all terms have the same degree) or normalize the variables (e.g., force a certain sum or product). In this case, since we already have a constraint $x+y+z=3$, we might not need to homogenize, but we should keep it in mind.
4. Symmetry: The cyclic nature of the inequality is a significant clue. Any technique we use should ideally respect this symmetry to make our lives easier. This means we should treat $x, y,$ and $z$ in a similar fashion throughout the proof.

Diving into the Proof: A Step-by-Step Approach

Let's start by trying to simplify the expression inside the square root. We have $x^2+yz+2$. Using the condition $x+y+z=3$, we can write $2$ as $2 \cdot 1 = 2(\frac{x+y+z}{3})$. This gives us:

$$x^2+yz+2 = x^2+yz+2(\frac{x+y+z}{3}) = x^2+yz+\frac{2}{3}(x+y+z)$$

This doesn't seem to simplify things directly. Instead, let's try a different approach. We can write $2$ as $2 \cdot (1^2)$. Since $1 = \frac{x+y+z}{3}$, we have:

$$2 = 2(\frac{x+y+z}{3})^2 = \frac{2}{9}(x+y+z)^2 = \frac{2}{9}(x^2+y^2+z^2+2xy+2yz+2zx)$$

So,

$$x^2+yz+2 = x^2+yz+\frac{2}{9}(x^2+y^2+z^2+2xy+2yz+2zx)$$

This looks even more complicated! It seems like directly substituting $x+y+z=3$ for 2 isn't the most fruitful path. Let's pivot and consider a different strategy – applying Cauchy-Schwarz Inequality.

Leveraging Cauchy-Schwarz Inequality: A Powerful Tool

The Cauchy-Schwarz Inequality is a cornerstone for solving inequalities, especially those involving sums and square roots. It comes in various forms, but a common one states that for real numbers $a_i$ and $b_i$:

$$(\sum a_i^2)(\sum b_i^2) \ge (\sum a_ib_i)^2$$

Or, in another form, which is more relevant to our problem:

$$(\sqrt{a_1}^2 + \sqrt{a_2}^2 + \sqrt{a_3}^2)(\sqrt{b_1}^2 + \sqrt{b_2}^2 + \sqrt{b_3}^2) \ge (\sqrt{a_1b_1} + \sqrt{a_2b_2} + \sqrt{a_3b_3})^2$$

However, a more pertinent form for our problem is the Engel form (also known as Titu's Lemma), which is a special case of Cauchy-Schwarz:

$$\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \frac{a_3^2}{b_3} \ge \frac{(a_1+a_2+a_3)^2}{b_1+b_2+b_3}$$

While this might be useful in some contexts, our problem involves sums of square roots, not sums of squares divided by something. So, let's circle back to the standard Cauchy-Schwarz Inequality. It feels like a more natural fit here.

Before directly applying Cauchy-Schwarz, let's try to massage the expression inside the square root a little more. We have $x^2+yz+2$. Notice that if we could somehow make this a perfect square, things would become much simpler. However, directly forcing a perfect square seems challenging. Let's try a subtler approach.

We want to show that $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$. Let's think about squaring both sides. Squaring sums of square roots can get messy quickly, but it might reveal some hidden structure. However, before we do that, let’s take a step back and see if we can bound each individual square root term.

Bounding Individual Terms: A Strategic Retreat

Sometimes, the key to solving an inequality is to find good bounds for individual terms. Let's focus on $\sqrt{x^2+yz+2}$. We want to find a lower bound for this expression. Since $x, y, z \ge 0$ and $x+y+z=3$, we know that $0 \le x, y, z \le 3$.

Consider the term $yz$. By AM-GM inequality, we have $\frac{y+z}{2} \ge \sqrt{yz}$, so $yz \le (\frac{y+z}{2})^2$. Also, since $y+z = 3-x$, we get $yz \le (\frac{3-x}{2})^2$. This might be useful.

Now, let's rewrite $x^2+yz+2$ as:

$$x^2+yz+2 = x^2+yz+2(\frac{x+y+z}{3}) = x^2+yz+\frac{2}{3}(x+y+z)$$

Still, it doesn't lead to an immediate simplification. Let's go back to our initial form $x^2 + yz + 2$. We know $yz \le (\frac{3-x}{2})^2$, so:

$$x^2+yz+2 \le x^2 + (\frac{3-x}{2})^2 + 2 = x^2 + \frac{9-6x+x^2}{4} + 2 = \frac{5x^2-6x+17}{4}$$

This gives us an upper bound for $x^2+yz+2$, but we need a lower bound for the square root. This approach doesn't seem to be directly helping us. Let's think about when the equality case might occur. Equality in our original inequality should happen when $x=y=z=1$. In this case, we have:

$$\sqrt{1^2+(1)(1)+2} + \sqrt{1^2+(1)(1)+2} + \sqrt{1^2+(1)(1)+2} = \sqrt{4} + \sqrt{4} + \sqrt{4} = 2+2+2 = 6$$

This confirms that the inequality holds for $x=y=z=1$. The fact that equality holds at $x=y=z=1$ gives us a crucial hint. It suggests that we should try to show that each term $\sqrt{x^2+yz+2}$ is greater than or equal to some expression involving $x$ that equals 2 when $x=1$. A linear bound might be a good starting point.

Seeking a Linear Bound: Connecting the Dots

Let's try to find a linear lower bound for $\sqrt{x^2+yz+2}$. We hypothesize that $\sqrt{x^2+yz+2} \ge ax+b$ for some constants $a$ and $b$. Since we want equality to hold at $x=y=z=1$, we need $\sqrt{1^2+(1)(1)+2} = 2 = a(1)+b$, so $a+b=2$. This gives us a relationship between $a$ and $b$.

We need to find suitable values for $a$ and $b$ such that the inequality holds. Let’s try $a=1$ and $b=1$. This would mean we want to prove $\sqrt{x^2+yz+2} \ge x+1$. Squaring both sides, we get:

$$x^2+yz+2 \ge (x+1)^2 = x^2+2x+1$$

This simplifies to $yz+2 \ge 2x+1$, or $yz+1 \ge 2x$. This is the inequality we need to prove.

Let's analyze this. We know $x+y+z=3$, so $y+z = 3-x$. By AM-GM, $yz \le (\frac{y+z}{2})^2 = (\frac{3-x}{2})^2$. So, we want to show:

$$(\frac{3-x}{2})^2 + 1 \ge 2x$$

$$\frac{9-6x+x^2}{4} + 1 \ge 2x$$

$$9-6x+x^2+4 \ge 8x$$

$$x^2-14x+13 \ge 0$$

$$(x-1)(x-13) \ge 0$$

This inequality holds when $x \le 1$ or $x \ge 13$. Since $0 \le x \le 3$, the inequality holds for $x \le 1$. However, it doesn't hold for all $x$ in $[0,3]$. So, $a=1$ and $b=1$ doesn't work.

Let's try a different approach. We know the function $f(x) = \sqrt{x^2+yz+2}$ is concave. We can apply Jensen's inequality. However, Jensen's inequality requires a concave function, and the sum of concave functions is concave. It seems a bit complicated here.

Back to Cauchy-Schwarz: A Fresh Perspective

Let's revisit Cauchy-Schwarz Inequality. We want to show $\sum\limits_{\mathrm{cyc}}\sqrt{x^2+yz+2}\ge 6$. We can apply Cauchy-Schwarz in the following form:

$$(\sum\limits_{\mathrm{cyc}} \sqrt{x^2+yz+2})^2 \ge 3(\sum (x^2+yz+2))$$

So, if we can show that $3(\sum (x^2+yz+2)) \ge 36$, then we are done. This simplifies to:

$$\sum (x^2+yz+2) \ge 12$$

$$x^2+y^2+z^2+xy+yz+zx+6 \ge 12$$

$$x^2+y^2+z^2+xy+yz+zx \ge 6$$

We know $(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx = 9$. Let $S = xy+yz+zx$. Then $x^2+y^2+z^2 = 9-2S$. Substituting this, we get:

$$9-2S+S \ge 6$$

$$9-S \ge 6$$

$$S \le 3$$

So, we need to prove that $xy+yz+zx \le 3$. We know $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) = 9$. Also, $x^2+y^2+z^2 \ge \frac{(x+y+z)^2}{3} = \frac{9}{3} = 3$. So,

$$9 = x^2+y^2+z^2+2(xy+yz+zx) \ge 3+2(xy+yz+zx)$$

$$6 \ge 2(xy+yz+zx)$$

$$3 \ge xy+yz+zx$$

Thus, $S = xy+yz+zx \le 3$ holds! We have successfully proven that $\sum\limits_{\mathrm{cyc}}\sqrt{x^2+yz+2}\ge 6$.

Conclusion: Triumph Through Tenacity

Wow, guys, we did it! We successfully navigated through a challenging inequality problem. By strategically simplifying the expression, considering various inequality techniques like Cauchy-Schwarz, bounding individual terms, and ultimately leveraging a clever application of Cauchy-Schwarz, we arrived at the solution. Remember, the key to solving these types of problems is persistence and a willingness to explore different approaches. Keep practicing, and you'll become inequality-solving ninjas in no time!

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Original Keyword: Prove $\sum\limits_{\mathrm{cyc}}\sqrt{x^2+yz+2}\ge 6$ for $x, y, z \ge 0$ with $x+y+z=3$

Fixed Keyword: How to prove the inequality $\sum\limits_{\mathrm{cyc}}\sqrt{x^2+yz+2}\ge 6$ given $x, y, z \ge 0$ and $x+y+z=3$?

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