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 20, 2025, 1:04 PM

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Let $x,y\ge 0$ such that $13(x^3+y^3) \leq 125(1+xy)$ . Prove that
$k(x+y)-xy\leq 5(2k-5)$ Where $k\geq 5.6797.$
$6(x+y)-xy\leq 35$

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42601 posts

sqing

#2 h Apr 20, 2025, 1:15 PM

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Let $x,y\ge 0$ such that $37(x^3+y^3) \leq 432(1+xy)$ . Prove that
$34(x+y)- 5xy\leq 228$ $7(x+y)-xy\leq 48$

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DAVROS

#3 h Apr 21, 2025, 9:17 AM

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sqing wrote:

Let $x,y\ge 0$ such that $13(x^3+y^3) \leq 125(1+xy)$ . Prove that
$k(x+y)-xy\leq 5(2k-5)$ Where $k\geq 5.6797.$

solution

This post has been edited 1 time. Last edited by DAVROS, Apr 21, 2025, 10:24 AM

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42601 posts

sqing

#4 h Apr 21, 2025, 1:15 PM

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Very very nice.Thank DAVROS:
Let $x,y\ge 0$ such that $13(x^3+y^3) \leq 125(1+xy)$ . Prove that
$k(x+y)-xy\leq 5(2k-5)$ Where $k\geq \frac{585}{103}.$

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42601 posts

sqing

#5 h Apr 21, 2025, 1:16 PM

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Let $x,y\ge 0$ such that $5(x^3+y^3) \leq 27(1+xy)$ . Prove that
$k(x+y)-xy\leq 3(2k-3)$ Where $k\geq 4.568.$
Let $x,y\ge 0$ such that $3(x^3+y^3) \leq 8(2+xy)$ . Prove that
$k(x+y)-xy\leq 4(k-1)$ Where $k\geq 2.4546.$

This post has been edited 1 time. Last edited by sqing, Apr 21, 2025, 1:21 PM

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sqing

#6 h Apr 25, 2025, 9:01 AM

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Let $a,b,c,d >0$ and $abcd=1$ . Prove that
$\frac{a+b+c}{ d+a+1 }+ \frac{b+c+d}{ a+b+1}+\frac{c+d+a}{b +c+1}+\frac{d+a+b}{c +d+1 } \geq4$

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#7 h Apr 25, 2025, 9:28 AM

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sqing wrote:

Let $a,b,c,d >0$ and $abcd=1$ . Prove that
$\frac{a+b+c}{ d+a+1 }+ \frac{b+c+d}{ a+b+1}+\frac{c+d+a}{b +c+1}+\frac{d+a+b}{c +d+1 } \geq4$

According to the Cauchy-Schwarz inequality,

$\sum \frac{a + b + c}{d + a + 1} \geq \frac{\left(\sum (a + b + c)\right)^2}{\sum (a + b + c)(d + a + 1)}$
Moreover,

$\left(\sum (a + b + c)\right)^2 - 4 \sum (a + b + c)(d + a + 1) = 3\left(\sum a - 4\right)\left(\sum a\right) + 2(a - c)^2 + 2(b - d)^2 \geq 0$

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sqing

#8 h Apr 25, 2025, 12:57 PM

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Very very nice.Thank lbh_qys.

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sqing

#9 h Apr 25, 2025, 1:02 PM

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Let $a,b,c,d >0$ and $abcd=1$ . Prove that
$\frac{a+b+c}{ d+2 }+ \frac{b+c+d}{ a+2}+\frac{c+d+a}{b +2 }+\frac{d+a+b}{c +2 } \geq4$

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42601 posts

sqing

#10 h Apr 26, 2025, 9:29 AM

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Let $x,y>0$ and $x+y+\frac{1}{x}+\frac{2}{y}=5.$ Prove that $\frac{13-\sqrt{41}}8\leq xy\leqslant\frac{13+\sqrt{41}}8$ kuing
Let $a,b,c>0$ and $a+b+c=3.$ Prove that $\frac 1a+\frac 1{ab}+\frac 1{abc}\geq \frac {3+\sqrt 5}2$

Attachments:

This post has been edited 1 time. Last edited by sqing, Apr 26, 2025, 12:19 PM

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sqing

#11 h Apr 27, 2025, 9:24 AM

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Let $a,b,c\geq 0$ and $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2 .$ Prove that
$ab+bc+ca+\frac{1}{4}\leq \frac{2}{3}\left(a+b+c\right)$ $ab+bc+ca+\frac{1}{3}\leq \frac{2}{3}\left(a+2b+c\right)$ $ab+2bc+ca \leq \frac{2}{3}\left(a+b+c+1\right)$ $ab+2bc+ca\leq\frac{2}{3}\left(a+2b+c+3-2\sqrt 2\right)$

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DAVROS

#12 h Apr 28, 2025, 3:07 PM

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sqing wrote:

Let $a,b,c\geq 0$ and $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2 .$ Prove that $ab+bc+ca+\frac{1}{3}\leq \frac{2}{3}\left(a+2b+c\right)$

solution

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sqing

#13 h Apr 28, 2025, 3:14 PM

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Very very nice.Thank DAVROS.

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