Inequalities

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sqing

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#1 h Apr 16, 2025, 4:52 AM • 1 Y

Y by Kizaruno

Let $a,b$ be reals such that $a^2+ab+b^2 =3$ . Prove that
$\frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$ $\frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$ $\frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$ Let $a,b$ be reals such that $a^2-ab+b^2 =1$ . Prove that
$\frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$ $\frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$ $\frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$

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#2 h Apr 17, 2025, 3:52 AM

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Let $a,b\geq 0$ and $a^2-ab+b^2 =1$ . Prove that
$\frac{5}{4} \geq\frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 } \geq \frac{1}{2 }$ $\frac{5}{6}\geq\frac{1}{ 2a^2+b+1 }+ \frac{1}{2b^2+a+1 } \geq \frac{1}{2 }$ $\frac{3}{2} \geq\frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 } +ab> \frac{6}{5}$ $\frac{3}{2} \geq\frac{1}{ 2a^2+b+1 }+ \frac{1}{2b^2+a+1 } +ab\geq \frac{5}{6}$ $\frac{5}{4} \geq \frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 }-ab\geq -\frac{1}{2 }$ $\frac{5}{6}\geq \frac{1}{ 2a^2+b+1 }+ \frac{1}{2b^2+a+1 }-ab\geq -\frac{1}{2 }$

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#3 h Apr 17, 2025, 4:26 AM • 1 Y

Y by Kizaruno

Let $a,b\geq 0$ and $a^2-ab+b^2 =3$ . Prove that
$\frac{53-\sqrt 3}{46} \geq\frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 } \geq \frac{7-\sqrt 3}{23}$ $\frac{7\sqrt 3-5}{14} \geq\frac{1}{ 2a^2+b+1 }+ \frac{1}{2b^2+a+1 } \geq \frac{7-\sqrt 3}{23}$ $\frac{76-\sqrt 3}{23} \geq\frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 } +ab\geq \frac{53-\sqrt 3}{46}$ $\frac{76-\sqrt 3}{23}\geq\frac{1}{ 2a^2+b+1 }+ \frac{1}{2b^2+a+1 } +ab\geq \frac{7\sqrt 3-5}{14}$ $\frac{53-\sqrt 3}{46} \geq \frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 }-ab\geq -\frac{62+\sqrt 3}{23 }$ $\frac{7\sqrt 3-5}{14}\geq \frac{1}{ 2a^2+b+1 }+ \frac{1}{2b^2+a+1 }-ab\geq -\frac{62+\sqrt 3}{23 }$

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#4 h Apr 17, 2025, 7:15 AM • 1 Y

Y by Kizaruno

sqing wrote:

Let $a,b$ be reals such that $a^2+ab+b^2 =3$ . Prove that $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$

solution

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#5 h Apr 17, 2025, 8:05 AM • 1 Y

Y by Kizaruno

Very very nice.Thank DAVROS.

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#6 h Apr 18, 2025, 1:48 AM

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Let $a,b\geq 0$ and $a+ b+2ab=4 .$ Prove that
$9(a+b-2)(11 - 10ab) \ge2\sqrt {29} (a-1)(b-1)(a-b)$ $9(a+b-2)(100- 99ab) \ge4\sqrt {74} (a-1)(b-1)(a-b)$ Let $a,b\geq 0$ and $2a+2 b+ab=5.$ Prove that
$15(a+b-2)(4- ab) \ge 8(a-1)(b-1)(a-b)$ $18(a+b-2)(6- 5ab)\ge\sqrt{119}(a-1)(b-1)(a-b)$ $18(a+b-2)(100- 99ab)\ge 5\sqrt{95}(a-1)(b-1)(a-b)$

This post has been edited 1 time. Last edited by sqing, Apr 18, 2025, 2:12 AM

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#7 h Apr 18, 2025, 2:23 AM

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Let $a,b\geq 0$ and $a+ b+ab+a^2+b^2=5.$ Prove that
$(9+\sqrt{21} ) (a+b-2)(7- 2ab)\ge14(a-1)(b-1)(a-b)$ $(9+\sqrt{21} ) (a+b-2)(4 - ab) \ge 8(a-1)(b-1)(a-b)$ Let $a,b\geq 0$ and $a+ b+ab=3 .$ Prove that
$8 (a+b-2)(6- 5ab) \ge \sqrt{39} (a-1)(b-1)(a-b)$ $8 (a+b-2)(100-99ab) \ge \sqrt{791} (a-1)(b-1)(a-b)$ Let $a,b\geq 0$ and $a+ b+2ab=4 .$ Prove that
$\sqrt{\frac{63\sqrt{21} -207}{1000}} (a^2+b^2+a+b-4) (3- 2ab) \ge (a-1)(b-1)(a-b)$

This post has been edited 3 times. Last edited by sqing, Apr 18, 2025, 3:33 AM

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#8 h Apr 18, 2025, 7:35 AM

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sqing wrote:

Let $a,b\geq 0$ and $a+ b+2ab=4 .$ Prove that $9(a+b-2)(11 - 10ab) \ge2\sqrt {29} (a-1)(b-1)(a-b)$

solution

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#9 h Apr 18, 2025, 8:39 AM

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sqing wrote:

Let $a,b$ be reals such that $a^2-ab+b^2 =1$ . Prove that $\frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$

solution

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#10 h Apr 18, 2025, 8:54 AM

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Very very nice.Thank DAVROS.

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#11 h Apr 19, 2025, 7:58 AM

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Let $a,b,c,d\geq0, (a+b)(c+d)=1 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq\frac{1}{2}$ Let $a,b,c,d\geq0, (a+2b)(c+2d)=1 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq\frac{1}{5}$ Let $a,b,c,d\geq0, (a+2b)(2c+ d)=1 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq\frac{3}{14}$

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#12 h Apr 19, 2025, 8:29 AM

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Let $a,b,c$ be reals such that $a+4b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq \frac{9}{2\sqrt[3]{4}}$ $a^2+2b^2+c^2\geq \frac{5}{ \sqrt[3]{4}}$

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#13 h Apr 19, 2025, 8:54 AM

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sqing wrote:

Let $a,b\geq 0$ and $a^2-ab+b^2 =3$ . Prove that $\frac{53-\sqrt 3}{46} \geq\frac{1}{ 2a^2+a+1 }+ \frac{1}{2b^2+b+1 } \geq \frac{7-\sqrt 3}{23}$

solution

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#14 h Apr 19, 2025, 9:35 AM

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sqing wrote:

Let $a,b,c,d\geq0, (a+b)(c+d)=1 .$ Prove that $a^2+b^2+c^2+d^2-ac-bd \geq\frac{1}{2}$

solution

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#15 h Apr 19, 2025, 12:11 PM

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Very very nice.Thank DAVROS.

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#16 h Apr 19, 2025, 12:42 PM

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Let $a,b\geq 0$ and $3a+3b+ab= 3$ . Prove that
$(a^3+1) (b^3+1)(a^3+b^3) \le2$ Let $a,b\geq 0$ and $a+b+ab= 3$ . Prove that
$(a^3+1) (b^3+1)(a^3+b^3) \le 756$

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#17 h Apr 19, 2025, 4:57 PM

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sqing wrote:

Let $a,b,c$ be reals such that $a+4b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that $a^2+b^2+c^2\geq \frac{9}{2\sqrt[3]{4}}$

solution

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#18 h Apr 19, 2025, 6:11 PM • 1 Y

Y by teomihai

sqing wrote:

Let $a,b\geq 0$ and $3a+3b+ab= 3$ . Prove that $(a^3+1) (b^3+1)(a^3+b^3) \le2$

solution

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#19 h Apr 20, 2025, 2:38 AM

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Very very nice.Thank DAVROS.

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#21 h Apr 20, 2025, 2:06 PM

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sqing wrote:

Let $x,y\ge 0$ such that $3(x+y)=1+xy.$ Prove that $x+9y+\frac{1}{x}+\frac{9}{y} \geq \frac{148}{3}$

solution

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#22 h Apr 20, 2025, 2:08 PM

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Very very nice.Thank DAVROS.
Let $a,b,c\geq 0$ and $ab+bc+ca=1.$ Prove that $\sqrt{\frac{a}{1+3ab}}+\sqrt{\frac{b}{1+3bc}}+\sqrt{\frac{c}{1+3ca}} \geq \sqrt 2$ https://artofproblemsolving.com/community/c6h386799p13607620

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#23 h Apr 20, 2025, 2:24 PM

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Let $x,y\ge 0$ such that $k(x+y)=1+xy.$ Prove that $x+k^2y+\frac{1}{x}+\frac{k^2}{y} \geq \frac{k^2(k+1)^2+(k-1)^2}{k}$ Where $k\in N^+.$

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#24 h Apr 21, 2025, 3:24 AM

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

This post has been edited 1 time. Last edited by sqing, Apr 21, 2025, 3:24 AM

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#25 h Apr 21, 2025, 3:35 AM

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

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#26 h Apr 21, 2025, 3:45 AM

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sqing wrote:

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

Note that $a, b, c \leq \sqrt{2}$ and $a + b + c \geq \sqrt{\sum ab + \sum a^2} = \sqrt{2}$ .

Therefore,

$\sum \left(1 - \frac{1}{\sqrt{a+1}}\right) = \sum \frac{\sqrt{a+1} - 1}{\sqrt{a+1}} = \sum \frac{a}{(\sqrt{a+1} + 1)\sqrt{a+1}} \geq \sum \frac{a}{(\sqrt{\sqrt{2}+1} + 1)\sqrt{\sqrt{2}+1}} \geq \frac{\sqrt{2}}{(\sqrt{\sqrt{2}+1} + 1)\sqrt{\sqrt{2}+1}} = 1 - \sqrt{\sqrt{2} - 1}.$