$$ \left\{\begin{array}{l}a+b+c=1 \a-\frac{1}{4} \b-\frac{1}{4} \c-\frac{1}{4} \\sqrt {4a+1}+\sqrt {4b+1}+\sqrt {4c+1}

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$$ \left\{\begin{array}{l}a+b+c=1 \a>-\frac{1}{4} \b>-\frac{1}{4} \c>-\frac{1}{4} \\sqrt {4a+1}+\sqrt {4b+1}+\sqrt {4c+1}<5\end{array}ight. $$

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Let's break down the problem step by step. We have the following system of equations and inequalities: 1. $$ a + b + c = 1 $$ 2. $$ a > -\frac{1}{4} $$ 3. $$ b > -\frac{1}{4} $$ 4. $$ c > -\frac{1}{4} $$ 5. $$ \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1} < 5 $$ ### Step 1: Analyze the first equation From the first equation, we can express one variable in terms of the others. Let's express $$ c $$: $$ c = 1 - a - b $$ ### Step 2: Substitute into the inequality Now, we substitute $$ c $$ into the inequality: $$ \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4(1 - a - b) + 1} < 5 $$ This simplifies to: $$ \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4 - 4a - 4b + 1} < 5 $$ $$ \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{5 - 4a - 4b} < 5 $$ ### Step 3: Define new variables Let $$ x = \sqrt{4a + 1} $$, $$ y = \sqrt{4b + 1} $$, and $$ z = \sqrt{5 - 4a - 4b} $$. From these definitions, we can express $$ a $$ and $$ b $$ in terms of $$ x $$ and $$ y $$: $$ a = \frac{x^2 - 1}{4}, \quad b = \frac{y^2 - 1}{4} $$ Substituting these into the equation for $$ c $$: $$ c = 1 - \frac{x^2 - 1}{4} - \frac{y^2 - 1}{4} = 1 - \frac{x^2 + y^2 - 2}{4} = \frac{6 - x^2 - y^2}{4} $$ ### Step 4: Substitute back into the inequality Now we need to ensure that: $$ x + y + \sqrt{5 - 4\left(\frac{x^2 - 1}{4} + \frac{y^2 - 1}{4}\right)} < 5 $$ This simplifies to: $$ x + y + \sqrt{5 - (x^2 + y^2 - 2)} < 5 $$ $$ x + y + \sqrt{7 - x^2 - y^2} < 5 $$ ### Step 5: Solve the inequality To solve this inequality, we can analyze the possible values of $$ x $$ and $$ y $$ under the constraints $$ a, b, c > -\frac{1}{4} $$. 1. Since $$ a > -\frac{1}{4} $$, we have $$ \sqrt{4a + 1} > 0 $$ which implies $$ x > 0 $$. 2. Similarly, $$ y > 0 $$. 3. The maximum value of $$ x + y $$ occurs when $$ z $$ is minimized. ### Step 6: Find the bounds To find the bounds for $$ x + y $$, we can set $$ z = 0 $$: $$ x + y < 5 $$ This gives us a region in the $$ xy $$-plane where $$ x + y < 5 $$ and both $$ x, y > 0 $$. ### Conclusion The solution set consists of all $$ (a, b, c) $$ such that: - $$ a + b + c = 1 $$ - $$ a, b, c > -\frac{1}{4} $$ - $$ \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1} < 5 $$ This defines a region in three-dimensional space, bounded by the conditions above. The exact values of $$ a, b, c $$ can be found by testing values within these constraints. The solution set includes all values of $$ a, b, c $$ that satisfy: 1. $$ a + b + c = 1 $$ 2. $$ a, b, c > -\frac{1}{4} $$ 3. $$ \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1} < 5 $$ These conditions define a region in three-dimensional space where the sum of $$ a, b, c $$ is 1, and each variable is greater than $$ -\frac{1}{4} $$, with the sum of their square roots being less than 5.

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