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sqing

#1 h Mar 19, 2025, 3:39 AM

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Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2 \ge 8$

This post has been edited 1 time. Last edited by sqing, Mar 19, 2025, 8:22 AM

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#2 h Mar 19, 2025, 6:59 AM

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sqing wrote:

Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2 \ge 8$

WLOG $a \geq b \geq c$ .

If $b > 2$ , then
$a^2 + b^2 + c^2 > 4 + 4 + 0 = 8.$
If $a < 2$ , then
$2\bigl(4 + abc - ab - bc - ca\bigr) = (2-a)(2-b)(2-c) + a(2-b)(2-c) + b(2-c)(2-a) + c(2-a)(2-b) > 0,$ a contradiction.

Therefore, it suffices to consider the case $a \geq 2 \geq b$ , in which
$8 \leq 2ab + 2c(a+b-ab) \leq 2ab + c^2 + (a+b-ab)^2 = a^2 + b^2 + c^2 + ab(a-2)(b-2) \leq a^2 + b^2 + c^2.$

This post has been edited 2 times. Last edited by lbh_qys, Mar 19, 2025, 7:00 AM

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#3 h Mar 19, 2025, 7:06 AM

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sqing wrote:

Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2-abc \ge 4$ $a^2 + b^2 + c^2+a^2b^2c^2\ge 8$ $a^2 + b^2 + c^2+ab+bc+ca \ge 12$

$a^2 + b^2 + c^2 - abc \ge ab+bc+ca-abc \ge 4$
$a^2 + b^2 + c^2 + a^2b^2c^2 \ge a^2 + b^2 + c^2 \ge 8$
$a^2 + b^2 + c^2 + ab + bc + ca \ge 8 + ab + bc + ca \ge 12$

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sqing

#4 h Mar 19, 2025, 8:21 AM

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Nice.Thank lbh_qys .Sorry.

This post has been edited 1 time. Last edited by sqing, Mar 19, 2025, 8:31 AM

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sqing

#5 h Mar 19, 2025, 8:30 AM

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Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2+a+b- ab\ge 8$ $a^2 + b^2 + c^2-ab \ge 4$ $a^2 + b^2 + c^2-\frac32 ab \ge 2$

This post has been edited 1 time. Last edited by sqing, Mar 19, 2025, 8:36 AM

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#6 h Mar 19, 2025, 8:44 AM

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sqing wrote:

Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2+a+b- ab\ge 8$

If $a+b-ab < 0$ , then $bc+ca \leq abc$ , hence $ab \geq 4$ .

Consequently, $a+b \geq 4$ and $a^2+b^2-ab \geq ab \geq 4$ .

If $a+b-ab \geq 0$ , then $a^2+b^2+c^2+a+b-ab \geq a^2+b^2+c^2 \geq 8$ .

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#7 h Mar 19, 2025, 8:45 AM

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sqing wrote:

Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2-ab \ge 4$

If $ab \ge 4$ , then $a^2+b^2-ab\ge ab\ge 4$ .

If $ab \le 4$ , then $a^2+b^2+c^2-ab\ge a^2+b^2+c^2-4\ge 4$ .

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#8 h Mar 19, 2025, 8:46 AM

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sqing wrote:

Let $a,b,c \ge 0$ and $ab + bc + ca \ge 4 + abc.$ Prove that
$a^2 + b^2 + c^2-\frac32 ab \ge 2$

If $ab \geq 4$ , then
$a^2 + b^2 - \frac{3}{2}ab \geq \frac{1}{2}ab \geq 2.$ If $ab \leq 4$ , then
$a^2 + b^2 + c^2 - \frac{3}{2}ab \geq a^2 + b^2 + c^2 - 6 \geq 2.$

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sqing

#9 h Mar 19, 2025, 8:47 AM

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Very very nice.Thank lbh_qys .

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