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

Grades 9-12, Ages 13-18

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sqing

#1 h Apr 9, 2025, 2:40 PM

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Let $a,b,c$ be real numbers so that $a+2b+3c=2$ and $2ab+6bc+3ca =1.$ Show that
$-\frac{1}{6} \leq ab-bc+ ca\leq \frac{1}{2}$ $\frac{5-\sqrt{61}}{9} \leq a-b+c\leq \frac{5+\sqrt{61}}{9}$

This post has been edited 1 time. Last edited by sqing, Apr 9, 2025, 2:41 PM

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sqing

#2 h Apr 10, 2025, 3:37 AM

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Let $a,b,c>0$ and $a+ 2b+c =2.$ Prove that
$\frac 1a + \frac 1{2b} + \frac 1c+abc \geq\frac{251}{54}$ Let $a,b,c>0$ and $2a+ b+2c = 2.$ Prove that
$\frac 1a + \frac 2b + \frac 1c+abc \geq\frac{245}{27}$

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#3 h Apr 10, 2025, 5:04 AM

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sqing wrote:

Let $a,b,c>0$ and $a+ 2b+c =2.$ Prove that
$\frac 1a + \frac 1{2b} + \frac 1c+abc \geq\frac{251}{54}$

According to the AM-GM inequality, we have
$2 = a + 2b + c\ge 3\sqrt[3]{2abc} \quad \Longrightarrow \quad abc \le \frac{4}{27}.$ Moreover,
$\frac{1}{a}+\frac{1}{2b}+\frac{1}{c}+abc \ge 3\sqrt[3]{\frac{1}{a}\cdot\frac{1}{2b}\cdot\frac{1}{c}}+abc = \frac{3}{(2abc)^{1/3}}+abc.$ The remaining task is to prove that when $0<abc\le\frac{4}{27}$ ,
$\frac{3}{(2abc)^{1/3}}+abc\ge \frac{251}{54},$ which is trivial.

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DAVROS

#4 h Apr 10, 2025, 6:54 AM

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sqing wrote:

Let $a,b,c$ be real numbers so that $a+2b+3c=2$ and $2ab+6bc+3ca =1.$ Show that $\frac{5-\sqrt{61}}{9} \leq a-b+c\leq \frac{5+\sqrt{61}}{9}$

solution

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sqing

#5 h Apr 10, 2025, 8:01 AM

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Very nice.Thank lbh_qys.

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sqing

#6 h Apr 10, 2025, 8:03 AM

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Very nice.Thank DAVROS.

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#7 h Apr 10, 2025, 8:33 AM

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sqing wrote:

Let $a,b,c$ be real numbers so that $a+2b+3c=2$ and $2ab+6bc+3ca =1.$ Show that
$\frac{5-\sqrt{61}}{9} \leq a-b+c\leq \frac{5+\sqrt{61}}{9}$

Let
$x = a - \frac{2}{3}, \quad y = 2b - \frac{2}{3}, \quad z = 3c - \frac{2}{3},$ then we have
$x + y + z = 0 \quad \text{and} \quad xy + yz + zx + 2(x+y+z) + \frac{12}{9} = 1.$ Thus,
$x+y+z = 0 \quad \text{and} \quad xy+yz+zx = -\frac{1}{3}.$ From this it follows that
$x^2 + y^2 + z^2 = \frac{2}{3}.$ Moreover,
$a - b + c = \frac{5}{9} + x - \frac{y}{2} + \frac{z}{3} = \frac{5}{9} + \frac{1-\frac{1}{2}+\frac{1}{3}}{3}(x+y+z) + \frac{13x-14y+z}{18} = \frac{5}{9} + \frac{13x-14y+z}{18}.$ According to the Cauchy–Schwarz inequality,
$(13x-14y+z)^2 \le (13^2+14^2+1^2)(x^2+y^2+z^2) = 244.$ Hence,
$\left| a-b+c-\frac{5}{9} \right| \le \frac{\sqrt{244}}{18} = \frac{\sqrt{61}}{9}.$

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sqing

#8 h Apr 10, 2025, 9:03 AM

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Very nice.Thank lbh_qys.

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sqing

#9 h Apr 11, 2025, 1:44 PM

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Let $a,b>0$ and $a^2+b^2 +ab+a+b=5$ . Prove that $\frac{1}{ a+2b }+ \frac{1}{ b+2a }+ \frac{1}{ab+2 } \geq 1$

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sqing

#10 h Apr 13, 2025, 10:17 AM

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Let $a,b\geq 0$ and $\frac{1}{a^2+b}+\frac{1}{b^2+a}=1.$ Prove that
$a^2+ab+b^2\leq 3$ $a^2-ab+b^2\leq \frac{3+\sqrt 5}{2}$ $a+b+a^3+b^3 \leq \frac{5+3\sqrt 5}{2}$ $a^2+b^2+a^3+b^3 \leq \frac{7+3\sqrt 5}{2}$

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sqing

#11 h Apr 13, 2025, 10:42 AM

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Let $a,b\geq 0$ and $\frac{a}{a^2+b}+\frac{b}{b^2+a}=1.$ Prove that
$a^2+b^2-ab\leq 1$ $a^2+b^2+ab\leq 3$ $a+b+a^3+b^3\leq 4$ $a^2+b^2+a^3+b^3\leq 4$

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DAVROS

#12 h Apr 14, 2025, 3:21 PM

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sqing wrote:

Let $a,b>0$ and $a^2+b^2 +ab+a+b=5$ . Prove that $\frac{1}{ a+2b }+ \frac{1}{ b+2a }+ \frac{1}{ab+2 } \geq 1$

solution

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sqing

#13 h Apr 15, 2025, 2:11 AM

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Very very nice.Thank DAVROS.

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sqing

#14 h Apr 17, 2025, 2:20 PM

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Let $a,b,c\geq 0$ and $ab+bc+ca=3.$ Prove that
$6(a+b+c-3)(11-5abc)\ge11(a-b)(b-c)(c-a)$ $6(a+b+c-3)(11-4abc)\ge 11(a-b)(b-c)(c-a)$ $2(a+b+c-3)(57-20abc)\ge 19(a-b)(b-c)(c-a)$ $6(a+b+c-3)(59-20abc)\ge 59(a-b)(b-c)(c-a)$

This post has been edited 2 times. Last edited by sqing, Apr 17, 2025, 2:59 PM

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