Inspired by Izho 2025

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Inspired by Izho 2025

G H J

Reveal topic tags

inequalities proposed

L

G H BBookmark kLocked kLocked NReply

Source: Own

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#1 h Mar 20, 2025, 2:42 AM

Y by

Let $a,b\geq 0$ and $(a - b)^2 + a + b \leq 2.$ Prove that
$1 \geq a^3 + b^3 - ab\geq -\frac{1}{27}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

425 posts

#2 h Mar 20, 2025, 2:50 AM

Y by

sqing wrote:

Let $a,b\geq 0$ and $(a - b)^2 + a + b \leq 2.$ Prove that
$1\geq a^3 + b^3 - ab\geq -\frac{1}{27}$

According to the AM-GM inequality, for any non-negative real numbers $a$ and $b$ , the inequality
$a^3 + b^3 + \frac{1}{27} \ge ab$ holds, and the left-hand side is almost indistinguishable from Izho 2025.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#3 h Mar 20, 2025, 2:59 AM

Y by

Let $a, b\geq 0$ and $(a-b)^2 + a + b \leq 2$ . Prove that
$-\frac{1}{25}<a^3 + b^3+ a^5+ b^5- ab \leq3$

This post has been edited 1 time. Last edited by sqing, Mar 20, 2025, 3:08 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#4 h Mar 20, 2025, 3:00 AM

Y by

Nice.Thank lbh_qys.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

425 posts

#5 h Mar 20, 2025, 3:05 AM

Y by

sqing wrote:

Let $a, b\geq 0$ and $(a-b)^2 + a + b \leq 2$ . Prove that
$\frac{1}{25}<a^3 + b^3+ a^5+ b^5- ab$

When $a = b = 0.3$ , $a^3 + b^3 + a^5 + b^5 - ab \approx -0.03 < 0$ ; do you mean $-\frac{1}{25}$ ?

This post has been edited 1 time. Last edited by lbh_qys, Mar 20, 2025, 3:05 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#6 h Mar 20, 2025, 3:06 AM

Y by

Let $a,b\geq 0$ and $(a - b)^2 + a + b \leq 2.$ Prove that
$2 \geq a^3 + b^3 +a^2b^2- ab\geq \frac{17\sqrt{17}-71}{32}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#7 h Mar 20, 2025, 3:08 AM

Y by

Yeah.Sorry.Thanks.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

425 posts

#8 h Mar 20, 2025, 3:10 AM

Y by

sqing wrote:

Let $a, b\geq 0$ . Prove that
$-\frac{1}{25}<a^3 + b^3+ a^5+ b^5- ab$

$a^3 + b^3 + a^5 + b^5 -ab \geq a^3+b^3-ab \geq -\frac{1}{27} > -\frac{1}{25}$
maybe $-\frac{1}{32}$

This post has been edited 2 times. Last edited by lbh_qys, Mar 20, 2025, 3:41 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#9 h Mar 20, 2025, 3:22 AM

Y by

very nice.Thank lbh_qys.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

425 posts

#10 h Mar 20, 2025, 3:44 AM

Y by

sqing wrote:

Let $a, b\geq 0$ and $(a-b)^2 + a + b \leq 2$ . Prove that
$a^3 + b^3+ a^5+ b^5- ab \leq3$

sqing wrote:

Let $a,b\geq 0$ and $(a - b)^2 + a + b \leq 2.$ Prove that
$2 \geq a^3 + b^3 +a^2b^2- ab$

When $(a-b)^2 + a + b \leq 2$ , then

$a^5 + b^5 \leq 2$ $a^2 b^2 \leq 1$ Thus, they can be deduced from $a^3 + b^3 - ab \leq 1$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41172 posts

sqing

#11 h Mar 20, 2025, 3:59 AM

Y by

very nice.Thank lbh_qys.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a