If $a,b,c$ are three positive real numbers, then $\frac{a^{2} + 1}{ ext{ }b + c} + \frac{b^{2} + 1}

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If $a,b,c$ are three positive real numbers, then $\frac{a^{2} + 1}{	ext{ }b + c} + \frac{b^{2} + 1}{c + a} + \frac{c^{2} + 1}{a + b}$ can never be equal to

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Sol. [Key Concept: Bound a symmetric expression in three positive reals (AM–GM).]$$\frac{a^{2} + 1}{b + c} + \frac{b^{2} + 1}{c + a} + \frac{c^{2} + 1}{a + b} \geq 2\left( \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} ight)$$$$\begin{matrix} & \ = 2\left\lbrack (a + b + c)\left( \frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} ight) - 3 ightbrack \geq 2\left( \frac{9}{2} - 3 ight) = 3 \ & \frac{3}{\left( \frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} ight)} \leq \frac{(b + c) + (c + a) + (a + b)}{3} \ & \ \Rightarrow (a + b + c)\left( \frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} ight) \geq \frac{9}{2} \end{matrix}$$✔ (a) expression $\geq 3$, so it can never equal 1✔ (b) can never equal 2 $( < 3)$✔ (c) can never equal $\frac{83\ ( < 3)}{}$✘ (d) equals 3 at $a = b = c$ — attainable

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