Inequalities from SXTX

Art of Problem Solving

AoPS Online

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

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

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

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

B

No tags match your search

M

Inequalities from SXTX

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

41401 posts

sqing

#1 h Feb 18, 2025, 12:25 PM

Y by

T702. Let $a,b,c>0$ and $a+2b+3c=\sqrt{13}.$ Prove that $\sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$ S
T703. Let $a,b$ be real numbers such that $a+b\neq 0.$ . Find the minimum of $a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $a,b,c>0$ and $a+b+c=3.$ Prove that $\frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)} \geq 6$ S

Attachments:

This post has been edited 5 times. Last edited by sqing, Feb 22, 2025, 3:48 AM

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#2 h Feb 18, 2025, 12:28 PM

Y by

Let $a,b,c$ be real numbers such that $a+b+c\neq 0.$ . Prove that $a^2+b^2+c^2 +\left( \frac{1-ab-bc-ca}{a+b+c}\right)^2 \geq \frac23$ Equality holds when $a=b=c=\frac{1}{\sqrt 6}$ or $a=b=c=-\frac{1}{\sqrt 6} .$
Let $a,b$ be real numbers such that $a+b\neq 0.$ . Prove that $a^2+b^2+\left(\frac{1-ab}{a+b} \right)^2\geq 1$ Equality holds when $a=b=\frac{1}{\sqrt 3}$ or $a=b=-\frac{1}{\sqrt 3} .$
h

Attachments:

This post has been edited 4 times. Last edited by sqing, Mar 15, 2025, 12:37 PM

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#3 h Feb 18, 2025, 2:06 PM

Y by

Let $a,b,c>0$ and $a+2b+3c= \sqrt{19}.$ Prove that $\sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq \sqrt{55}$

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#4 h Feb 19, 2025, 2:39 PM

Y by

sqing wrote:

Let $a,b$ be real numbers such that $a+b\neq 0.$ . Prove that $a^2+b^2+\left(\frac{1-ab}{a+b} \right)^2\geq 1$ Equality holds when $a=b=\frac{1}{\sqrt 3}$ or $a=b=-\frac{1}{\sqrt 3} .$

$a^2+b^2+\left(\frac{1-ab}{a+b} \right)^2\geq\frac{ a^2+b^2}{2}+ab+\left(\frac{1-ab}{a+b} \right)^2\geq\frac{(a+b)^2}{4}+ab+\left(\frac{1-ab}{a+b} \right)^2$ $\geq 2\sqrt{\frac{(a+b)^2}{4} \cdot \left(\frac{1- ab}{a+ b}\right)^2}+ab = |1-ab| +ab\geq 1$ h

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#5 h Mar 13, 2025, 2:36 PM

Y by

T706. Let $x ,y$ be real number . Find the minimum of $|2x^2+3x-2-2y^2|+ |2x^2-3x-2-2y^2|.$
T707. Let $x>0.$ Find the maximum of $\frac{x^2}{\sqrt{3x^4+1}} + 2\sqrt{\frac{x}{3x^2+1} }.$
T708. Let $a,b,c>0$ be real numbers. Prove that
$\frac a{bc(1+c)}+\frac b{ca(1+a)}+\frac c{ab(1+a)}\geq \frac 3{ 1+abc}$ T709. Let $a,b,c>0$ and $\frac{1}{a+2} +\frac{1}{b+2} +\frac{1}{c+2} =1.$ Prove that $\sqrt{ab(4-ab)} + \sqrt{bc(4-bc)} +\sqrt{ca(4-ca)} \leq \frac{\sqrt{3(8+abc)^3}}{9}$

Attachments:

This post has been edited 3 times. Last edited by sqing, Mar 15, 2025, 8:29 AM

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#6 h Mar 13, 2025, 2:39 PM

Y by

Let $x ,y$ be real number . Prove that $|2x^2+3x-2-2y^2|+ |2x^2-3x-2-2y^2|\geq 3$ Let $x>0.$ Prove that
$\frac{x^2}{\sqrt{3x^4+1}} + 2\sqrt{\frac{x}{3x^2+1} } \leq \frac 32$

This post has been edited 1 time. Last edited by sqing, Mar 13, 2025, 2:40 PM

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#8 h Mar 15, 2025, 8:29 AM

Y by

sqing wrote:

Let $x ,y$ be real number . Prove that $|2x^2+3x-2-2y^2|+ |2x^2-3x-2-2y^2|\geq 3$ Let $x>0.$ Prove that
$\frac{x^2}{\sqrt{3x^4+1}} + 2\sqrt{\frac{x}{3x^2+1} } \leq \frac 32$

Attachments:

This post has been edited 1 time. Last edited by sqing, Mar 16, 2025, 1:22 AM

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#9 h Mar 17, 2025, 3:19 AM

Y by

Let $x>0.$ Prove that
$\frac{x+1}{\sqrt{3x^2+1}} + \frac{2x^2}{3x^2+1} \leq \frac 32$ $\frac{x+1}{\sqrt{3x^2+1}} + \frac{ x }{3x^2+1} \leq \frac{1}{\sqrt[3]2}+ \frac{1}{\sqrt[3]4}$

This post has been edited 1 time. Last edited by sqing, Mar 17, 2025, 3:21 AM

Z K Y

The post below has been deleted. Click to close.

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

476 posts

#10 h Mar 17, 2025, 4:54 AM

Y by

sqing wrote:

Let $x>0.$ Prove that
$\frac{x+1}{\sqrt{3x^2+1}} + \frac{2x^2}{3x^2+1} \leq \frac 32$

Let
$u=\frac{x}{\sqrt{3x^2+1}},\qquad v=\frac{1}{\sqrt{3x^2+1}},$ then
$3u^2 + v^2 = 1.$
Hence,
$\begin{aligned} \text{RHS} - \text{LHS} &= \frac{3}{2} - (u+v) - 2u^2 \\ &=\frac{1}{2} + (1-2u^2) - (u+v) \\ &=\frac{1}{2} + (u^2+v^2) - (u+v) \\ &=\left(u-\frac{1}{2}\right)^2+\left(v-\frac{1}{2}\right)^2 \ge 0. \end{aligned}$

Z K Y

The post below has been deleted. Click to close.

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

476 posts

#11 h Mar 17, 2025, 6:34 AM

Y by

sqing wrote:

Let $x>0.$ Prove that
$\frac{x+1}{\sqrt{3x^2+1}} + \frac{ x }{3x^2+1} \leq \frac{1}{\sqrt[3]2}+ \frac{1}{\sqrt[3]4}$

Let $u = \frac{x}{\sqrt{3x^2+1}}, v = \frac{1}{\sqrt{3x^2+1}}, t = \sqrt[3]{2}$ , then
$2(t^2-1)(\mathrm{RHS}-\mathrm{LHS}) = (1-3u^2-v^2) + 3\left(u - \frac{t^2-1}{3} (v+1)\right)^2 + \frac{t}{t+1} (tv-1)^2 \ge 0$

This post has been edited 1 time. Last edited by lbh_qys, Mar 17, 2025, 6:46 AM

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#12 h Mar 17, 2025, 7:30 AM

Y by

Very very nice.Thank lbh_qys.
Let $x>0.$ Prove that
$\frac{2x^2}{3x^2+1}+ \sqrt{\frac{2}{x+1}} \leq \frac 32$

Attachments:

This post has been edited 1 time. Last edited by sqing, Mar 20, 2025, 8:40 AM

Z K Y

The post below has been deleted. Click to close.

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

4 posts

#13 h Mar 22, 2025, 3:47 PM

Y by

good job

Z K Y

The post below has been deleted. Click to close.

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

41401 posts

sqing

#14 h Mar 30, 2025, 2:07 PM

Y by

Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $abcd \geq 9.$ Prove that
$-6\leq ad+bc \leq 6$ $\frac{7}{2}\leq a^2+bc+d^2 \leq \frac{19}{2}$ $-\frac{13}{2}\leq ab+cd \leq \frac{13}{2}$ Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $a+bc+d \leq -5 .$ Prove that
$abcd\geq -9$ $-6\leq ad+bc \leq 6$ $7\leq a+b^2+c^2+d^2 \leq \frac{53}{4}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a