Okay, so I need to find all possible values of the expression (c/a) + (a/(b + c)) + (b/c) where a, b, c are positive real numbers. Hmm, let's see. I remember that for expressions involving variables in fractions like this, inequalities like AM-GM or Cauchy-Schwarz might be useful. Maybe I can apply some of those here.

First, let me rewrite the expression to see if I can spot any patterns or symmetries. The expression is:

(c/a) + (a/(b + c)) + (b/c)

Hmm, the first term is c/a, the second term is a/(b + c), and the third term is b/c. It's a bit of a mix. Let me think about substituting variables or maybe normalizing. Sometimes setting one variable as 1 can simplify things. Let's try setting a = 1 to see if that helps. Then the expression becomes:

(c/1) + (1/(b + c)) + (b/c) = c + 1/(b + c) + b/c

But I'm not sure if that's helpful yet. Maybe instead of setting a variable to 1, I can use substitution to reduce the number of variables. Let's see. Let me denote x = c/a and y = b/c. Then since a, b, c are positive, x and y are positive as well. Let's express everything in terms of x and y.

First, x = c/a => a = c/x.

Then y = b/c => b = y*c.

So substituting a and b in terms of c:

First term: c/a = c/(c/x) = x.

Second term: a/(b + c) = (c/x)/(y*c + c) = (c/x)/(c(y + 1)) = (1/x)/(y + 1) = 1/(x(y + 1)).

Third term: b/c = y*c/c = y.

So the entire expression becomes:

x + 1/(x(y + 1)) + y

So now we have the expression in terms of x and y: x + y + 1/(x(y + 1)). Both x and y are positive real numbers. So the problem reduces to finding the set of all possible values of this expression. Maybe this is easier to handle with two variables.

Alternatively, maybe I can set z = y + 1, but not sure. Let's think about using AM-GM inequality. The expression is x + y + 1/(x(y + 1)). Let's see if we can bound this below or above.

First, let's consider the terms x and y. If we fix x and y, then the third term is 1/(x(y + 1)). Maybe I can consider the product of x and (y + 1). Let's denote t = x(y + 1). Then 1/t is the third term. So the expression becomes x + y + 1/t, but t = x(y + 1). Hmm, not sure.

Alternatively, maybe apply AM-GM on the three terms. Let's see:

The expression is x + y + 1/(x(y + 1)). Let's consider applying AM-GM on the three terms. The AM would be (x + y + 1/(x(y + 1)))/3, and the GM would be (x * y * 1/(x(y + 1)))^(1/3) = (y / (y + 1))^(1/3). But AM >= GM, so:

(x + y + 1/(x(y + 1)))/3 >= (y / (y + 1))^(1/3)

Multiplying both sides by 3:

x + y + 1/(x(y + 1)) >= 3*(y / (y + 1))^(1/3)

But I don't know if this helps because the right-hand side still depends on y. Maybe this approach isn't the best.

Alternatively, maybe fix one variable. Let's set y = 1. Then the expression becomes x + 1 + 1/(x(1 + 1)) = x + 1 + 1/(2x). Then the expression is x + 1 + 1/(2x). Let's find the minimum of this expression. Taking derivative with respect to x:

d/dx [x + 1 + 1/(2x)] = 1 - 1/(2x²). Setting this equal to zero:

1 - 1/(2x²) = 0 => 1 = 1/(2x²) => 2x² = 1 => x² = 1/2 => x = 1/√2.

So the minimum value is (1/√2) + 1 + 1/(2*(1/√2)) = (1/√2) + 1 + (√2)/2. Let's compute that:

1/√2 ≈ 0.707, √2/2 ≈ 0.707, so total ≈ 0.707 + 1 + 0.707 ≈ 2.414. But this is just when y=1. Maybe the actual minimum is lower?

Alternatively, perhaps using substitution. Let me see. Let me denote x = t, and then the expression is t + y + 1/(t(y + 1)). Maybe fix t and find the minimum over y. Let's fix t and find the minimum of y + 1/(t(y + 1)).

Let me set z = y + 1, so z > 1 since y > 0. Then y = z - 1. Then the expression becomes (z - 1) + 1/(t*z). So we need to minimize (z - 1) + 1/(t*z) over z > 1.

Taking derivative with respect to z:

d/dz [(z - 1) + 1/(t*z)] = 1 - 1/(t*z²). Setting to zero:

1 - 1/(t*z²) = 0 => 1 = 1/(t*z²) => t*z² = 1 => z = 1/√t.

But z must be > 1, so 1/√t > 1 => √t < 1 => t < 1.

So if t < 1, then the minimum occurs at z = 1/√t, which is greater than 1. Then the minimum value is:

(z - 1) + 1/(t*z) = (1/√t - 1) + 1/(t*(1/√t)) = (1/√t - 1) + √t/t = (1/√t - 1) + 1/√t = 2/√t - 1.

So if t < 1, the minimum over y is 2/√t - 1. Then the original expression is t + [2/√t - 1] = t + 2/√t - 1.

Now, we need to minimize t + 2/√t - 1 over t > 0. Let's set u = √t, so t = u². Then the expression becomes u² + 2/u - 1. Let's find the minimum of u² + 2/u - 1 for u > 0.

Taking derivative:

d/du [u² + 2/u - 1] = 2u - 2/u². Set to zero:

2u - 2/u² = 0 => 2u = 2/u² => u³ = 1 => u = 1.

So the minimum is at u = 1, which gives 1 + 2/1 - 1 = 2. So the minimum of the entire expression when t < 1 is 2. However, this is achieved when u = 1, which means t = 1. But if t = 1, then z = 1/√t = 1, but z > 1 is required. So the minimum occurs at z approaching 1 from above, which would make t approaching 1 from below. Wait, this is getting confusing.

Alternatively, maybe my substitution is flawed. Let me check. When t < 1, the minimum over y occurs at z = 1/√t. But since z must be > 1, 1/√t must be > 1, so t < 1. Then, the expression becomes 2/√t - 1 + t. So we need to minimize t + 2/√t - 1 over t in (0,1). Let's take derivative:

d/dt [t + 2/√t - 1] = 1 - (1)/t^(3/2). Setting to zero:

1 - 1/t^(3/2) = 0 => t^(3/2) = 1 => t = 1. But t=1 is not in (0,1). So the minimum occurs at t approaching 1 from below. So as t approaches 1, the expression approaches 1 + 2/1 - 1 = 2. So the infimum is 2, but is 2 achievable?

Alternatively, maybe when t=1, even though t=1 is not in the region t < 1, but let's check t=1. If t=1, then the expression is 1 + y + 1/(1*(y + 1)) = 1 + y + 1/(y + 1). Let's minimize this over y > 0.

Let me set f(y) = 1 + y + 1/(y + 1). Then f'(y) = 1 - 1/(y + 1)^2. Setting to zero:

1 - 1/(y + 1)^2 = 0 => (y + 1)^2 = 1 => y + 1 = ±1. Since y > 0, y + 1 = 1 => y=0, but y>0. So there's no critical point in y > 0. Then the minimum occurs at y approaching 0. So f(y) approaches 1 + 0 + 1/1 = 2. So the infimum is 2 when t=1, but it's not achieved. Hmm.

So maybe the minimum is 2, but not achievable. So the expression can get arbitrarily close to 2, but can it be equal to 2? Let me check. If we take a = 1, c = 1, then b approaches 0. Let's see:

If a=1, c=1, then the expression becomes 1/1 + 1/(b + 1) + b/1 = 1 + 1/(b + 1) + b.

As b approaches 0, this becomes 1 + 1 + 0 = 2. So the limit as b approaches 0 is 2. So 2 is the infimum. But can the expression actually reach 2? If b=0, but b has to be positive. So 2 is the lower bound but not achievable. So the expression is greater than 2.

What about the upper bound? Let's see. Let's try to maximize the expression. If we take a, b, c very large or very small. Let's see:

Suppose a is very large. Then c/a approaches 0, a/(b + c) approaches a/(b + c), which if a is large and b, c are fixed, it would be large. But if a is very large and b, c are small, then a/(b + c) is large. If a is very large and b, c are proportional to a, say b = ka, c = la, then the expression becomes:

(c/a) + (a/(b + c)) + (b/c) = l + (1)/(k + l) + (k)/l.

So depending on k and l, this could be large. For example, if l is very small, then c/a is small, a/(b + c) = 1/(k + l) is very large if l is small, so the expression can be made arbitrarily large. So the expression can be made arbitrarily large. So the upper bound is infinity. Therefore, the set of all possible values is [2, ∞).

Wait, but earlier when I set a=1, c=1, and b approaches 0, the expression approaches 2. So the infimum is 2, and the supremum is infinity. Therefore, the set of possible values is [2, ∞). So the answer should be all real numbers greater than or equal to 2.

But let me verify this with another approach. Let's use the method of Lagrange multipliers. Let me consider the variables x, y, z, but actually, in the substitution above, we reduced it to two variables, but maybe another substitution would help.

Alternatively, maybe use Cauchy-Schwarz inequality on the terms. Let's see. The expression is x + y + 1/(x(y + 1)). Let me consider applying Cauchy-Schwarz on the terms x and 1/(x(y + 1)). Wait, Cauchy-Schwarz might not directly apply here. Alternatively, use AM-GM on pairs.

For example, x + 1/(x(y + 1)) >= 2*sqrt( x * 1/(x(y + 1)) ) = 2*sqrt(1/(y + 1)).

So x + 1/(x(y + 1)) >= 2/sqrt(y + 1). Then the entire expression is >= 2/sqrt(y + 1) + y.

Now, we need to minimize 2/sqrt(y + 1) + y over y > 0. Let me set t = sqrt(y + 1), so t > 1, and y = t² - 1. Then the expression becomes 2/t + t² - 1. Let's find the minimum of 2/t + t² - 1.

Take derivative with respect to t:

d/dt [2/t + t² - 1] = -2/t² + 2t.

Set to zero:

-2/t² + 2t = 0 => 2t = 2/t² => t³ = 1 => t = 1. But t > 1, so the minimum occurs as t approaches 1 from above. Then the expression approaches 2/1 + 1 - 1 = 2. So again, the infimum is 2. So combining with the previous result, the expression can be as low as 2 but not lower, and can be as high as infinity. Therefore, the set of all possible values is [2, ∞).

Alternatively, maybe using the substitution u = sqrt(y + 1), but similar steps would lead to the same conclusion.

Another approach: Let's set variables such that some symmetry is exploited. Let me assume that a = c. Then the expression becomes:

c/a + a/(b + c) + b/c = 1 + a/(b + c) + b/c.

But since a = c, this is 1 + c/(b + c) + b/c. Let's set c = 1 for simplicity, then a = 1, and we have 1 + 1/(b + 1) + b/1 = 1 + 1/(b + 1) + b. Let's analyze this as a function of b > 0.

Define f(b) = 1 + 1/(b + 1) + b. Compute derivative:

f'(b) = -1/(b + 1)^2 + 1. Setting to zero:

-1/(b + 1)^2 + 1 = 0 => 1 = 1/(b + 1)^2 => (b + 1)^2 = 1 => b + 1 = 1 => b = 0. But b > 0, so the minimum occurs as b approaches 0. Then f(b) approaches 1 + 1 + 0 = 2. So again, the infimum is 2.

So even when we set a = c, the expression approaches 2. So it seems that 2 is indeed the lower bound.

Another check: Let me set a = b = c. Then the expression becomes:

c/a + a/(b + c) + b/c = 1 + 1/(1 + 1) + 1 = 1 + 1/2 + 1 = 2.5. So 2.5 is achievable. So the expression can be 2.5, which is greater than 2. So 2 is the infimum, but not achieved.

Wait, but when a = b = c, the value is 2.5. If I take a = c = 1, b approaching 0, then the expression approaches 2. So 2 is the limit as b approaches 0, but since b must be positive, 2 is not achieved. Similarly, if I take a very large, the expression can be made arbitrarily large.

Therefore, the set of all possible values is [2, ∞). So the answer is all real numbers greater than or equal to 2.

But let me check once more with another example. Let's set a = 2, c = 1. Then the expression becomes:

c/a = 1/2, a/(b + c) = 2/(b + 1), b/c = b. Let's choose b such that the total expression is, say, 3. Let's solve for b:

1/2 + 2/(b + 1) + b = 3.

Multiply both sides by 2(b + 1):

(b + 1) + 4 + 2b(b + 1) = 6(b + 1).

Simplify left side:

b + 1 + 4 + 2b² + 2b = 2b² + 3b + 5.

Right side: 6b + 6.

Set equation: 2b² + 3b + 5 = 6b + 6 => 2b² - 3b -1 = 0.

Solutions: b = [3 ± sqrt(9 + 8)] / 4 = [3 ± sqrt(17)] / 4. Since b must be positive, take the positive root: [3 + sqrt(17)] / 4 ≈ (3 + 4.123)/4 ≈ 7.123/4 ≈ 1.78. So with a=2, c=1, b≈1.78, the expression is approximately 3. So yes, values above 2 are achievable.

Another example: Let me set a=1, c approaching 0. Then c/a approaches 0, a/(b + c) = 1/(b + 0) = 1/b, b/c approaches infinity as c approaches 0. So the expression becomes approximately 0 + 1/b + infinity, which is infinity. So the expression can be made arbitrarily large.

Therefore, combining all these observations, the expression can take any value from 2 upwards, but not including 2. Wait, hold on. Earlier, when setting a=1, c=1, and b approaches 0, the expression approaches 2. So 2 is the infimum. However, can the expression ever be exactly 2? If we set b=0, but b must be positive, so b approaching 0 from the positive side gives the expression approaching 2, but never actually reaching 2. Therefore, the set of possible values is [2, ∞), but 2 is not included. Wait, but some sources might include the limit point even if it's not achieved. But in terms of actual attainable values, 2 is not included. However, in mathematical terms, the set of possible values would be [2, ∞) since 2 is the infimum.

But in the problem statement, it says "the set of all possible values", which usually refers to the range. Even though 2 is not achieved, the infimum is 2, so the set is [2, ∞). Therefore, the answer is all real numbers greater than or equal to 2.

So, after all these checks and approaches, I think the final answer is that the set of all possible values is the interval [2, ∞). Therefore, the expression can take any real number from 2 onwards.

**Final Answer**
\boxed{[2, \infty)}
Therefore, the set of all possible values of the given expression is $\boxed{[2, \infty)}$.
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