MHB What are the real numbers a, b, and c that satisfy certain conditions?

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To find the real numbers a, b, and c that meet the conditions a^2 + b^2 + c^2 = 26, a + b = 5, and b + c ≥ 7, the equations can be manipulated to express c in terms of a and b. Substituting b = 5 - a into the first equation leads to a quadratic in terms of a. The inequality b + c ≥ 7 can also be expressed using the derived values of b and c. Solving these equations reveals the possible values for a, b, and c that satisfy all conditions. The solution process involves algebraic manipulation and consideration of inequalities to ensure all criteria are met.
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Determine all three real numbers $a, \,b$ and $c$ which satisfies the conditions $a^2+ b^2+ c^2= 26$, $a + b = 5$ and $b + c\ge 7.$
 
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My attempt:

$a^2+b^2+c^2 = 26\;\;\;\;(1).$

$a+b = 5\;\;\;\;(2).$

$b+c \ge 7\;\;\;\;(3).$

Combining $(1)$ and $(2)$:

\[a^2+b^2+c^2 = (5-b)^2+b^2+c^2 = 26\Rightarrow 2b^2-10b+c^2-1 = 0, \;\;\;\;(4).\]

From $(3)$ one gets: \[c^2 \geq (7-b)^2, \;\;\;\; (5).\]

Combining $(4)$ and $(5)$:

\[2b^2-10b+c^2-1 = 0 \geq 2b^2-10b+(7-b)^2-1 = 3b^2-24b+48\]

\[\Rightarrow b^2-8b+16 \leq 0\]

Or \[(b-4)^2 \leq 0\]

So the only possible $b$-value is $4$. Thus $a = 1$ (from $(2)$) and $c = 3$ (from $(1)$ and $(3)$).
 
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