Inequality while on a trip

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giangtruong13

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giangtruong13

#1 h Apr 12, 2025, 12:24 PM • 1 Y

Y by cubres

I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$ . Find the maximum: $A= \sum_{cyc} \frac{1}{16+a^3}$

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sqing

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sqing

#2 h Apr 12, 2025, 1:47 PM • 2 Y

Y by cubres, arqady

Let $a,b,c \geq -2$ such that $a^2+b^2+c^2 \leq 8.$ Prove that $\frac{1}{16+a^3}+\frac{1}{16+b^3}+\frac{1}{16+c^3}\leq \frac{5}{16}$ pretty fun

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giangtruong13

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giangtruong13

#3 h Apr 14, 2025, 4:12 PM • 1 Y

Y by cubres

Bummer tummer

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no_room_for_error

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no_room_for_error

#4 h Apr 14, 2025, 6:53 PM

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giangtruong13 wrote:

I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$ . Find the maximum: $A= \sum_{cyc} \frac{1}{16+a^3}$

$\frac{1}{16+a^3}=-\frac{a^2(a+2)(a^2-2a+8)}{64(a^3+16)}+\frac{1}{64}a^2+\frac{1}{16}\leq \frac{1}{64}a^2+\frac{1}{16}$

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GeoMorocco

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GeoMorocco

#5 h Apr 14, 2025, 7:23 PM

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giangtruong13 wrote:

I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$ . Find the maximum: $A= \sum_{cyc} \frac{1}{16+a^3}$

Obviously:
$\sum_{cyc} \frac{1}{16+a^3} \leq \sum_{cyc} \frac{1}{16-|a|^3}$ so we only need to study the inequality for negative numbers and the function is convex for $-2 \leq x \leq 0$ .

The function is convex with a convex constraint, therefore it is enough to check the boundaries and we get a maximum at $(0,-2,-2)$ and the maximum value is equal to $\frac{5}{16}$ .

This post has been edited 2 times. Last edited by GeoMorocco, Apr 16, 2025, 7:53 AM

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arqady

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arqady

#6 h Apr 16, 2025, 6:46 AM

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GeoMorocco wrote:

giangtruong13 wrote:

I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$ . Find the maximum: $A= \sum_{cyc} \frac{1}{16+a^3}$

The function is convex...

$\left( \frac{1}{16+a^3}\right)''=\frac{12a(a^3-8)}{(a^3+16)^3}$

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GeoMorocco

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GeoMorocco

#7 h Apr 16, 2025, 7:51 AM

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arqady wrote:

$\left( \frac{1}{16+a^3}\right)''=\frac{12a(a^3-8)}{(a^3+16)^3}$

Obviously:
$\sum_{cyc} \frac{1}{16+a^3} \leq \sum_{cyc} \frac{1}{16-|a|^3}$ so we only need to study the inequality for negative numbers and the function is convex for $-2 \leq x \leq 0$ .

This post has been edited 1 time. Last edited by GeoMorocco, Apr 16, 2025, 7:52 AM

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arqady

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arqady

#8 h Apr 16, 2025, 10:05 AM

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GeoMorocco wrote:

arqady wrote:

$\left( \frac{1}{16+a^3}\right)''=\frac{12a(a^3-8)}{(a^3+16)^3}$

Obviously:
$\sum_{cyc} \frac{1}{16+a^3} \leq \sum_{cyc} \frac{1}{16-|a|^3}$ so we only need to study the inequality for negative numbers and the function is convex for $-2 \leq x \leq 0$ .

Can you write a new conditions and an inequality, that we need to prove now?

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GeoMorocco

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GeoMorocco

#9 h Apr 16, 2025, 1:11 PM

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arqady wrote:

Can you write a new conditions and an inequality, that we need to prove now?

if at least one of the variables $a,b,c>0$ , we get:
$\sum_{cyc} \frac{1}{16+a^3} < \frac{1}{16+0^3}+ \frac{1}{16+(-2)^3}+\frac{1}{16+(-2)^3}=\frac{1}{16}+\frac{1}{8}+\frac{1}{8}=\frac{5}{16}$ Therefore, it is enough to study the function for $a,b,c\leq 0$ . But the function $f(x)=\frac{1}{16+x^3}$ is convex for $-2\leq x\leq 0$ with convex constraints, so it is enough to check the borders where we get a maximum at $(0,-2,-2)$ equal to $\frac{5}{16}$ .

This post has been edited 1 time. Last edited by GeoMorocco, Apr 16, 2025, 1:19 PM

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arqady

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arqady

#10 h Apr 17, 2025, 5:50 AM

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GeoMorocco wrote:

arqady wrote:

Can you write a new conditions and an inequality, that we need to prove now?

if at least one of the variables $a,b,c>0$ , we get:
$\sum_{cyc} \frac{1}{16+a^3} < \frac{1}{16+0^3}+ \frac{1}{16+(-2)^3}+\frac{1}{16+(-2)^3}=\frac{1}{16}+\frac{1}{8}+\frac{1}{8}=\frac{5}{16}$ Therefore, it is enough to study the function for $a,b,c\leq 0$ . But the function $f(x)=\frac{1}{16+x^3}$ is convex for $-2\leq x\leq 0$ with convex constraints, so it is enough to check the borders where we get a maximum at $(0,-2,-2)$ equal to $\frac{5}{16}$ .

For $(-2,-2,-2)$ we obtain a greater value. Why don't we get a greater value with a condition $a^2+b^2+c^2\leq8$ ?

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GeoMorocco

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GeoMorocco

#11 h Apr 17, 2025, 7:59 AM

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arqady wrote:

For $(-2,-2,-2)$ we obtain a greater value. Why don't we get a greater value with a condition $a^2+b^2+c^2\leq8$ ?

You can't choose $(-2,-2,-2)$ as it does not verify the condition $a^2+b^2+c^2 \leq 8$ !!!! I know you are a very smart guy :first:

, but not sure what do you mean with your suggestion.

Here is the new problem after we proved that it is not optimal to have a positive coordinate:

Let $a,b,c$ be real numbers such as $-2 \leq a,b,c \leq 0$ and $a^2+b^2+c^2 \leq 8$ . Prove that:
$\sum_{cyc} \frac{1}{16+a^3} \leq \frac{5}{16}$

This post has been edited 2 times. Last edited by GeoMorocco, Apr 17, 2025, 8:24 AM

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arqady

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arqady

#12 h Apr 17, 2025, 8:30 AM • 1 Y

Y by teomihai

GeoMorocco wrote:

arqady wrote:

For $(-2,-2,-2)$ we obtain a greater value. Why don't we get a greater value with a condition $a^2+b^2+c^2\leq8$ ?

You can't choose $(-2,-2,-2)$ as it does not verify the condition $a^2+b^2+c^2 \leq 8$ !!!!

Read please better my previous post.
The function $f$ is convex on $[-2,0]$ and without condition $a^2+b^2+c^2\leq8$ you can get a maximal value by checking $\{a,b,c\}\subset\{-2,0\}$ .
Why with this condition you still can take $\{a,b,c\}\subset\{-2,0\}$ ? If for any $\{a,b,c\}\subset\{-2,0\}$ you'll obtain that the condition indeed occurs, so your reasoning is true. Otherwise, your reasoning is total wrong, I think.

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GeoMorocco

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GeoMorocco

#13 h Apr 17, 2025, 8:52 AM

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arqady wrote:

Read please better my previous post.
The function $f$ is convex on $[-2,0]$ and without condition $a^2+b^2+c^2\leq8$ you can get a maximal value by checking $\{a,b,c\}\subset\{-2,0\}$ .
Why with this condition you still can take $\{a,b,c\}\subset\{-2,0\}$ ? If for any $\{a,b,c\}\subset\{-2,0\}$ you'll obtain that the condition indeed occurs, so your reasoning is true. Otherwise, your reasoning is total wrong, I think.

I still don't get your point. The condition $a^2+b^2+c^2 \leq 8$ is essential to finding $a,b,c$ not just the border conditions. Check my example:
https://artofproblemsolving.com/community/c6t243f6h3550230_find_the_maxium_of_the_following_expression

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arqady

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arqady

#14 h Apr 17, 2025, 9:04 AM

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GeoMorocco, in your linked problem a maximal value occurs also for $a=-1$ and $-1\notin\{-2,0\}$ .
I hope, now you understood me.

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GeoMorocco

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GeoMorocco

#15 h Apr 17, 2025, 9:14 AM

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arqady wrote:

GeoMorocco, in your linked problem a maximal value occurs also for $a=-1$ and $-1\notin\{-2,0\}$ .
I hope, now you understood me.

Yes, of course. if you check my proof earlier, I said: "convex constraints" and not "convex constraint" which means that I used both constraints to find the maximum.

GeoMorocco wrote:

Therefore, it is enough to study the function for $a,b,c\leq 0$ . But the function $f(x)=\frac{1}{16+x^3}$ is convex for $-2\leq x\leq 0$ with convex constraints, so it is enough to check the borders where we get a maximum at $(0,-2,-2)$ equal to $\frac{5}{16}$ .

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