Inequality

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

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#1 h Mar 19, 2025, 3:13 AM

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$\begin{array}{l} a,b,c \ge 1,ab + bc + ca \ge 4 + abc\\ {a^2} + {b^2} + {c^2} \ge 9? \end{array}$
I tried:
$\begin{array}{l} * {\text{In three numbers }}a,b,c \ge 1,{\text{ at least 2 numbers }} \ge {\text{2 or}} \le {\text{2}}\\ {\text{WLOG assume they are }}a{\text{ and }}b \Rightarrow \left( {a - 2} \right)\left( {b - 2} \right) \ge 0 \Leftrightarrow 4c + abc \ge 2ac + 2bc\\ \Leftrightarrow 2ac + 2bc \le 4c - 4 + 4 + abc \le 4c - 4 + ab + bc + ca \Leftrightarrow ac + bc \le 4c - 4 + ab \end{array}$
How do I process further?

This post has been edited 3 times. Last edited by BaCaPhe, Mar 19, 2025, 3:24 AM

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#2 h Mar 19, 2025, 4:43 AM • 2 Y

Y by MihaiT, BaCaPhe

Let
$p = (a-1)+(b-1)+(c-1), \quad q = (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1), \quad r=(a-1)(b-1)(c-1).$ Then, the condition can be written as
$q + 2p + 3 \geq 4 + r + q + p + 1,$ i.e.,
$p \geq r + 2,$ which implies
$p \geq 2.$
Based on Schur's inequality,
$r \geq \frac{(4q-p^2)p}{9},$ it follows that
$p \geq r + 2 \geq \frac{(4q-p^2)p}{9} + 2,$ and hence
$q \leq \frac{p^3+9p-18}{4p}.$
This shows that
$a^2 + b^2 + c^2 - 9 = p^2+2p-2q-6 \geq p^2+2p-\frac{p^3+9p-18}{2p}-6 = \frac{(p-2)(p^2+6p-9)}{2p} \geq 0.$

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