hard problem

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Cobedangiu

70 posts

Cobedangiu

#1 h Mar 27, 2025, 2:54 PM • 1 Y

Y by PikaPika999

problem

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Cobedangiu

70 posts

Cobedangiu

#2 h Mar 27, 2025, 4:22 PM • 1 Y

Y by PikaPika999

no one? .

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Alex-131

5382 posts

Alex-131

#3 h Mar 27, 2025, 4:41 PM • 1 Y

Y by PikaPika999

Wrong solution in-english class, will check out later

This post has been edited 5 times. Last edited by Alex-131, Mar 27, 2025, 4:54 PM
Reason: r

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ND_

51 posts

ND_

#4 h Mar 27, 2025, 4:45 PM • 1 Y

Y by PikaPika999

When is equality achieved? a=b=c doesn't satisfy the initial condition. I think minimum is 10.

This post has been edited 2 times. Last edited by ND_, Mar 27, 2025, 4:50 PM

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Nuran2010

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Nuran2010

#7 h Mar 27, 2025, 4:50 PM • 2 Y

Y by PikaPika999, TunarHasanzade

https://artofproblemsolving.com/community/c6h1834400p23600351
Also posted here

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Cobedangiu

70 posts

Cobedangiu

#8 h Mar 27, 2025, 4:53 PM • 1 Y

Y by PikaPika999

ND_ wrote:

When is equality achieved? a=b=c doesn't satisfy the initial condition. I think minimum is 10.

with equality occurring when $a=2, b=1,c=1$

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ND_

51 posts

ND_

#10 h Mar 27, 2025, 5:04 PM • 1 Y

Y by PikaPika999

$b+c \ge 2\sqrt{bc} \iff \frac{b+c}{bc} \ge \frac{4}{b+c} \iff$ $\frac{1}{b} + \frac{1}{c} \ge \frac{2}{b+c} + \frac{2}{b+c}$
$\Rightarrow (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge \left(a+\frac{b+c}{2}+\frac{b+c}{2}\right)\left(\frac{1}{a} + \frac{2}{b+c} + \frac{2}{b+c}\right)$

Hence, minimum occurs at $b=c$ . Plugging onto original equation, we get $2 + \left(\frac{b}{a}\right)^3=5\left(\frac{b}{a}\right)$ , or $b=2a$ . So, minimum of 10 occurs at $(2b, b, b)$ .

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EaZ_Shadow

1267 posts

EaZ_Shadow

#11 h Mar 27, 2025, 5:07 PM • 1 Y

Y by PikaPika999

I used Cauchy and got $(a+b+c)(\frac1a+\frac1b+\frac1c)\geq9$ so i think 10 is the min, and I dont think 9 is achievable.

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xytunghoanh

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xytunghoanh

#12 h Mar 27, 2025, 5:09 PM • 1 Y

Y by PikaPika999

Cobedangiu wrote:

problem

cmath?

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Cobedangiu

70 posts

Cobedangiu

#13 h Mar 27, 2025, 5:17 PM • 2 Y

Y by xytunghoanh, PikaPika999

xytunghoanh wrote:

Cobedangiu wrote:

problem

cmath?

yes, but this inequality is from someone else, just took it back from that person.

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xytunghoanh

33 posts

xytunghoanh

#14 h Mar 27, 2025, 5:20 PM • 1 Y

Y by PikaPika999

I can help you to solve this. Check pm for Vietnamese solution.

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Cobedangiu

70 posts

Cobedangiu

#15 h Mar 27, 2025, 5:26 PM • 2 Y

Y by xytunghoanh, PikaPika999

xytunghoanh wrote:

I can help you to solve this. Check pm for Vietnamese solution.

I know how to do it

This post has been edited 1 time. Last edited by Cobedangiu, Mar 27, 2025, 5:27 PM

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sqing

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sqing

#16 h Mar 28, 2025, 3:03 AM • 1 Y

Y by PikaPika999

Cobedangiu wrote:

problem

Let $a,b,c>0$ and $a^3+b^3+c^3=5abc.$ Prove that $10 \leq (a+b+c)\left(\frac {1}{a}+\frac {1}{b}+\frac {1}{c}\right)\leq5+4\sqrt{2}$ https://artofproblemsolving.com/community/c6h3455063p33365029

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InftyByond

207 posts

InftyByond

#17 h Apr 5, 2025, 11:07 PM • 2 Y

Y by Maximilian113, PikaPika999

k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ to get $\sum (1+\frac{a}{b}+\frac{b}{a})$ into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ .
this now gives us $\sum (1+x+\frac{1}{x})$ which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$

and then going back into the condition
we should get $a=2, b=1, c=1$ must be the minimum and consequently the minimum value is achieved at 10????????????

This post has been edited 1 time. Last edited by InftyByond, Apr 6, 2025, 5:34 AM
Reason: fake proof issues die hard

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Cobedangiu

70 posts

Cobedangiu

#18 h Apr 6, 2025, 9:57 AM • 1 Y

Y by PikaPika999

InftyByond wrote:

k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ to get $\sum (1+\frac{a}{b}+\frac{b}{a})$ into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ .
this now gives us $\sum (1+x+\frac{1}{x})$ which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$

and then going back into the condition
we should get $a=2, b=1, c=1$ must be the minimum and consequently the minimum value is achieved at 10????????????

The problem has been solved

, but the solution will not be posted here

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Nguyenhuyen_AG

3322 posts

Nguyenhuyen_AG

#19 h Apr 6, 2025, 10:43 AM • 3 Y

Y by PikaPika999, anduran, truongphatt2668

Cobedangiu wrote:

problem

See here: https://nguyenhuyenag.wordpress.com/2025/04/06/inequality-43/

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InftyByond

207 posts

InftyByond

#20 h Apr 6, 2025, 4:33 PM • 1 Y

Y by PikaPika999

Cobedangiu wrote:

InftyByond wrote:

k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ to get $\sum (1+\frac{a}{b}+\frac{b}{a})$ into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ .
this now gives us $\sum (1+x+\frac{1}{x})$ which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$

and then going back into the condition
we should get $a=2, b=1, c=1$ must be the minimum and consequently the minimum value is achieved at 10????????????

The problem has been solved

, but the solution will not be posted here

K so i looked at the solve
Its pretty nice and slick with no calculus
Is my alternate solve wrong then? not surprised

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DKI

32 posts

DKI

#21 h Apr 7, 2025, 5:35 AM

Y by

we have
$(a+b-c)(a+b-2c+2\sqrt2c)(a+b-2c-2\sqrt2c)=4(a^3+b^3+c^3-5abc)-(a-b)^2(3a+3b+5c)$ hence $a+b\ge c$ , and similarly $a+c\ge b$ , $b+c\ge a$ ,
$T=(a+b+c)(\frac1a+\frac1b+\frac1c)=10+\frac{a^3+b^3+c^3-5abc+(a+b-c)(a+c-b)(b+c-a)}{abc}\ge10$

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giangtruong13

144 posts

giangtruong13

#22 h Apr 15, 2025, 7:47 AM

Y by

See this https://artofproblemsolving.com/community/c6h1834400p23632179 for more information

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