Inequalities

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

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

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sqing

#1 h Apr 22, 2025, 5:05 AM • 1 Y

Y by toanrathay

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1.$ Prove that $|a-b|+|b-2c|+|c-3a|\leq 5$ $|a-2b|+|b-3c|+|c-4a|\leq \sqrt{42}$ $|a-b|+|b-\frac{11}{10}c|+|c-a|\leq \frac{29}{10}$

This post has been edited 2 times. Last edited by sqing, Apr 22, 2025, 7:40 AM

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sqing

#2 h Apr 22, 2025, 8:50 AM

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Let $a,b,c$ be real numbers such that $|a-2b|+|b-2c|+|c-2a|=\sqrt{19}.$ Prove that $1\leq a^2+b^2+c^2\leq \frac{57}{7}$ Let $a,b,c$ be real numbers such that $|a-2b|+|b-c|+|c-2a|= 3\sqrt{2}.$ Prove that $1\leq a^2+b^2+c^2\leq 42$ Let $a,b,c$ be real numbers such that $|a-3b|+|b-c|+|c-3a|= 4\sqrt{2}.$ Prove that $1\leq a^2+b^2+c^2\leq \frac{91}{2}$

This post has been edited 1 time. Last edited by sqing, Apr 22, 2025, 8:54 AM

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#3 h Apr 22, 2025, 11:16 AM

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For the moment only the first one. Mod $p\rightarrow -p$ we may assume $a\ge b$ , then $\phi:=a-b+d(b-2c)+e(c-3a)=(1-3e)a+(d-1)b+(e-2d)c$ where $d,e\in \{-1,1\}$ .
Hence $\phi^2\le (1-3e)^2+(d-1)^2+(e-2d)^2$ . Checking the four cases gives a maximum 25 when $d=1,e=-1$ . Hence $\phi\le 5$ and we have equality when $(a,b,c)=1/5(4,0,-3)$ .

The other two problems are analogous. I checked.

This post has been edited 1 time. Last edited by alexheinis, Apr 22, 2025, 6:39 PM

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sqing

#4 h Apr 22, 2025, 1:06 PM

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Thanks.

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#5 h Apr 22, 2025, 1:09 PM

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sqing wrote:

Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1.$ Prove that $|a-b|+|b-2c|+|c-3a|\leq 5$

Equality holds when $a=-\frac{4}{5},b=0,c=\frac{3}{5}$ or $a= \frac{4}{5},b=0,c=-\frac{3}{5}.$

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sqing

#6 h Apr 24, 2025, 4:00 AM • 1 Y

Y by Havu

Let $a,b,c \ge \frac{3}{10}$ and $a^2+b^2+c^2=3$ .Prove that
$\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}\geq 1$

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sqing

#8 h Apr 25, 2025, 1:38 PM

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Let $a,b\geq 0$ and $a^2+b^2=1 .$ Prove that
$\frac{a}{ 1+ab+b }+\frac{b }{ 1+ab+a }+\frac{ab }{\frac{1}{3}+ ab }\leq \frac{6\sqrt 2 }{7 }+\frac{1 }{35 }$ $\frac{a}{ 1+ab+b }+\frac{b }{ 1+ab+a }+\frac{ab }{1+\frac{1}{3} ab }\leq \frac{6\sqrt 2-1}{7}$

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