Inequalities

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sqing

#1 h Jun 14, 2025, 6:47 AM

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Let $a,b\geq 0,a+b+ab=3 .$ Prove that
$\frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2} \geq \frac{3}{2}$ $\frac{10}{3}\geq \frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2}+ab \geq \frac{5}{2}$ $\frac{a}{ab+1}+\frac{b }{b+1}+\frac{1 }{a+1} \geq \frac{3}{2}$ $\frac{13}{4} \geq\frac{a}{ab+1}+\frac{b }{b+1}+\frac{1 }{a+1}+ ab\geq \frac{7}{4}$

This post has been edited 3 times. Last edited by sqing, Jun 14, 2025, 9:10 AM

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sqing

#2 h Jun 14, 2025, 7:40 AM

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

This post has been edited 1 time. Last edited by sqing, Jun 14, 2025, 8:14 AM

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#3 h Jun 14, 2025, 8:33 AM

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sqing wrote:

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

From the assumption we get $2 \leq a+b \leq 3$ . Let us prove a more general problem:
Let $a,b,c \geq 0, bc+ca+ab=3, 3 \leq a+b+c \leq 4$ , prove that:
$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2} \geq \frac{3}{2}$ Let $x=a+b+c \in [3;4]$ . Applying the AM-GM and Cauchy-Schwarz inequalities, we have:
$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}=\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2} \geq \frac{(a+b+c)^2}{ab+ac^2+bc+ba^2+ca+cb^2}$ Since $bc+ca+ab=3$ and $ac^2+ba^2+cb^2 \leq \frac{1}{9}(a+b+c)^3$ , we just need to prove:
$\frac{9x^2}{x^3+27} \geq \frac{3}{2} \Longleftrightarrow 18x^3 \geq 3x^3+81 \Longleftrightarrow (x-3)(9+3x-x^2) \geq 0$ Obviously the last inequality is true because $3 \leq x \leq 4$ . Equality at $a=b=c=1$ .

This post has been edited 2 times. Last edited by pooh123, Jun 14, 2025, 8:35 AM

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sqing

#4 h Jun 14, 2025, 8:43 AM

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Very very nice.Thank pooh123.

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#5 h Jun 14, 2025, 12:54 PM

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sqing wrote:

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

For $a,b,c \geq 0,bc+ca+ab=3$ , we'll prove that:
$\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a} \geq 3$ Let $x=a+b+c \geq 3$ and applying the Cauchy-Schwarz inequality, we have:
$\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a}=\frac{(a^2+b)(1+b)}{(a+b)(1+b)}+\frac{(b^2+c)(1+c)}{(b+c)(1+c)}+\frac{(c^2+a)(1+a)}{(c+a)(1+a)}$ $\geq \frac{(a+b)^2}{(1+b)(a+b)}+\frac{(b+c)^2}{(1+c)(c+a)}+\frac{(c+a)^2}{(1+a)(a+b)} \geq \frac{4(a+b+c)^2}{a^2+b^2+c^2+bc+ca+ab+2a+2b+2c}$ Since $a^2+b^2+c^2+bc+ca+ab=(a+b+c)^2-bc-ca-ab=(a+b+c)^2-3$ , we just need to prove:
$\frac{4x^2}{x^2+2x-3} \geq 3 \Longleftrightarrow 4x^2 \geq 3x^2+6x-9 \Longleftrightarrow (x-3)^2 \geq 0$ Obviously the last inequality is true. Equality at $a=b=c=1$ .

This post has been edited 1 time. Last edited by pooh123, Jun 14, 2025, 12:54 PM

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sqing

#6 h Jun 15, 2025, 6:43 AM

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Very nice.Thank pooh123:
Let $a,b,c\geq 0,ab+bc+ca =3 .$ Prove that
$\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a} \geq 3$ Let $a,b,c\geq 0,a+b+c=3 .$ Prove that
$\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a} \geq 3$

This post has been edited 1 time. Last edited by sqing, Jun 15, 2025, 6:44 AM

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sqing

#7 h Jun 15, 2025, 8:04 AM

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Let $a,b,c>0.$ Prove that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{3}{2}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2}\sqrt[3]{9}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{3}{4}\sqrt{\frac{2c}{a+b}+1}\geq \frac{3}{2}\sqrt[3]{\frac{9}{4}}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{1}{4}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2\sqrt[3]{4}}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{5}{2}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2}\sqrt[3]{25}$

This post has been edited 1 time. Last edited by sqing, Jun 15, 2025, 9:31 AM

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sqing

#8 h Jun 15, 2025, 8:13 AM

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Let $a,b,c>0.$ Prove that
$\frac{a}{b+c}+\frac{b}{c+a}+2\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq3\sqrt[3]{3}$ $\frac{a}{b+c}+\frac{b}{c+a}+3\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq\frac{9}{\sqrt[3]{4}}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{1}{2}\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq \frac{3}{2}\sqrt[3]{\frac{3}{2}}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{1}{3}\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq\sqrt[3]{\frac{9}{4}}$

This post has been edited 1 time. Last edited by sqing, Jun 15, 2025, 8:14 AM

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DAVROS

#9 h Jun 15, 2025, 9:09 AM

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sqing wrote:

Let $a,b,c>0.$ Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{3}{2}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2}\sqrt[3]{4}$

solution

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sqing

#10 h Jun 15, 2025, 9:30 AM

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Very very nice.Thank DAVROS.Sorry.

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sqing

#11 h Jun 15, 2025, 12:45 PM

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Let $a,b>0,a+b-ab=1.$ Prove that
$\dfrac{10}{3} \geq \dfrac{a}{b+1}+\dfrac{b}{a+1}+ab-\sqrt{2-ab} \geq 1-\sqrt 2$ Let $a,b>0,a+b+ab=3.$ Prove that
$3-\sqrt 2 \geq \dfrac{a}{b+1}+\dfrac{b}{a+1}+ab-\sqrt{2-ab} \geq 1$

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#12 h Jun 16, 2025, 1:25 AM

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sqing wrote:

Let $a,b>0,a+b-ab=1.$ Prove that
$\dfrac{10}{3} \geq \dfrac{a}{b+1}+\dfrac{b}{a+1}+ab-\sqrt{2-ab} \geq 1-\sqrt 2$

From $a+b-ab=1$ , we get $(a-1)(b-1)=0$ so one of $a,b$ is equal to $1$ .
WLOG, let $b=1$ , then $0 \leq a \leq 2$ and the inequality reduces to proving:
$\frac{10}{3} \geq \frac{3a}{2}+\frac{1}{a+1}-\sqrt{2-a} \geq 1-\sqrt{2}$ Minimum: We have:
$\frac{3a}{2}+\frac{1}{a+1}-\sqrt{2-a} \geq 1-\sqrt{2} \Longleftrightarrow \frac{3a}{2}+\frac{1}{a+1}-1 \geq \sqrt{2-a}-\sqrt{2}$ $\Longleftrightarrow \frac{a(3a+1)}{2(a+1)} \geq \frac{-a}{\sqrt{2-a}+\sqrt{2}} \Longleftrightarrow \frac{a(3a+1)}{2(a+1)}+\frac{a}{\sqrt{2-a}+\sqrt{2}} \geq 0$ The last inequality is true because $0 \leq a \leq 2$ . Equality at $a=0,b=1$ .
Maximum: We have:
$\frac{3a}{2}+\frac{1}{a+1}-\sqrt{2-a} \leq \frac{10}{3} \Longleftrightarrow \sqrt{2-a} \geq \frac{3a}{2}+\frac{1}{a+1}-\frac{10}{3}$ $\Longleftrightarrow \sqrt{2-a} \geq \frac{(a-2)(9a+7)}{6(a+1)} \Longleftrightarrow \sqrt{2-a}+\frac{(2-a)(9a+7)}{6(a+1)} \geq 0$ The last inequality is true because $0 \leq a \leq 2$ . Equality at $a=2,b=1$ .

This post has been edited 1 time. Last edited by pooh123, Jun 16, 2025, 1:26 AM

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sqing

#13 h Jun 16, 2025, 2:54 AM

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Very very nice.Thank pooh123.

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sqing

#14 h Jun 17, 2025, 4:03 AM

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Let $a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 .$ Prove that
$\frac{17}{16} \geq \frac{ a^5+b^5+c^5}{abc(a+b+c)^2} \geq \frac{13}{20}$ $\frac{ a^5+b^5+c^5}{(ab+bc+ca)(a+b+c)^3} \geq \frac{13}{200}$ $\frac{ a^5+b^5+c^5}{(a^2+b^2+c^2)(a+b+c)^3} \geq \frac{13}{225}$

This post has been edited 2 times. Last edited by sqing, Jun 18, 2025, 7:55 AM

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sqing

#15 h Jun 17, 2025, 5:28 AM

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Let $a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 .$ Prove that
$\frac{971+21\sqrt{21}}{2000}\geq\frac{\frac{3}{2}a^2+ b^2+\frac{3}{2}c^2}{(a+b+c)^2}\geq \frac{7}{16}$ $\frac{9}{25}> \frac{\frac{12}{5}a^3+ b^3+\frac{12}{5}c^3}{(a+b+c)^3}\geq \frac{1}{5}$

This post has been edited 1 time. Last edited by sqing, Jun 18, 2025, 7:59 AM

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DAVROS

#16 h Jun 18, 2025, 5:39 AM

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sqing wrote:

Let $a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 .$ Prove that $\frac{ a^5+b^5+c^5}{abc(a+b+c)^2} \geq \frac{13}{20}$

solution

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sqing

#17 h Jun 18, 2025, 6:57 AM

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Very very nice.Thank DAVROS:
Let $a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 .$ Prove that $\frac{17}{16} \geq \frac{ a^5+b^5+c^5}{abc(a+b+c)^2} \geq \frac{13}{20}$

This post has been edited 1 time. Last edited by sqing, Jun 18, 2025, 7:07 AM

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DAVROS

#18 h Jun 18, 2025, 7:31 AM

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sqing wrote:

Let $a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 .$ Prove that $\frac{ a^5+b^5+c^5}{(a^2+b^2+c^2)(a+b+c)^3} \geq \frac{13}{225}$

solution

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sqing

#19 h Jun 18, 2025, 8:05 AM

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Very very nice.Thank DAVROS.

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sqing

#20 h Jun 19, 2025, 2:06 PM

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Let $a,b,c \ge 0, a^2 + b^2 + c^2=1$ . Prove that
$\frac {1}{2 - a(b+c)} + \frac {1}{2 - b(c+a)} + \frac {1}{2 - c(a+b)} \leq \frac{9}{4}$

This post has been edited 1 time. Last edited by sqing, Jun 19, 2025, 2:09 PM

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sqing

#21 h Jun 21, 2025, 2:57 AM

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Let $a,b,c> 0 .$ Prove that $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{a^2+b^2+2c(a+b)}{ab+bc+ ca}+1$ $\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\geq \frac{a^2+b^2+2c(a+b)}{ab+bc+ ca}+1$

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