Inequalities

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sqing

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sqing

#1 h Mar 10, 2025, 2:21 AM

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Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that
$(a^2-a+1)(b^2-b+1) \geq 9$ $(a^2-a+b+1)(b^2-b+a+1) \geq 25$ Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=\frac{2}{3}.$ Prove that
$(a+8)(a^2-a+b+2)(b^2-b+5)\geq1331$ $(a+10)(a^2-a+b+4)(b^2-b+7)\geq2197$

This post has been edited 2 times. Last edited by sqing, Mar 10, 2025, 3:15 AM

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#2 h Mar 10, 2025, 2:28 AM

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Let $a,b,c>0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$(3a-1)( b-1)(3c-1) \geq 120$ $(3a-1)( 3b-1)(3c-1) \geq 512$ $(2a-1)(3b-1)(2c-1)\geq 99+45\sqrt5$ $(3a-1)( 2b-1)(3c-1)\geq157+26\sqrt{39}$

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DAVROS

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#3 h Mar 10, 2025, 7:49 AM

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sqing wrote:

Let $a,b,c>0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that $(3a-1)( 2b-1)(3c-1)\geq157+26\sqrt{39}$

solution

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#4 h Mar 10, 2025, 8:20 AM

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Very very nice.Thank DAVROS.
Let $a,b,c,d\geq 0$ and $a+b+c+d=1.$ Prove that
$\dfrac{a}{4b^2+1}+\dfrac{b}{4c^2+1}+\dfrac{c}{4d^2+1}+\dfrac{d}{4a^2+1}\geqslant \dfrac{3}{4}$ K
Let $a,b>0 .$ Prove that $(a^4+1)( b^4+1)+4ab\geq 2(ab+1)(a^2+b^2)$ $(a^6+1)( b^6+1)+4ab\geq 2(ab+1)(a^3+b^3)$

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#5 h Mar 10, 2025, 9:24 AM

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Let $a, b\geq 0$ and $a+b+7\leq3\sqrt{2a+2b+5}.$ Prove that
$a+3b+2ab\leq \frac{13}{2}$ $3a+2b+ab\leq \frac{25}{4}$ $4a+3b+ 2ab\leq \frac{73}{8}$ $2a+3b+4ab\leq \frac{145}{16}$

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#6 h Mar 10, 2025, 11:17 AM

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sqing wrote:

Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that
$(a^2-a+1)(b^2-b+1) \geq 9$

Solution

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#7 h Mar 10, 2025, 12:20 PM

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Very nice.Thanks.
Let $a,b\geq 2 .$ Prove that
$(1-a^2)(1-b^2) -2ab\geq 1$ $(1-a^3)(1-b^3) -3a^2b^2\geq 1$ $(1-a^2)(1-b^2) (1-ab)+7ab\leq 1$ $(1-a^3)(1-b^3) (1-ab)+37ab\leq 1$ Let $a,b,c\geq 2 .$ Prove that
$(a^2-1)(b^2-1)(c^2-1) -3abc\geq 3$ $(a^3-1)(b^3-1)(c^3-1) -5a^2b^2c^2\geq 23$ Let $a,b,c\geq 1 .$ Prove that
$(5-\frac{2a^2}{b^3})(5-\frac{2b^2}{c^3})(5-\frac{2c^2}{a^3})\leq 27a^2b^2c^2$

This post has been edited 2 times. Last edited by sqing, Mar 19, 2025, 5:19 AM

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#8 h Mar 10, 2025, 12:49 PM

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Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that
$(a^2-2a+2)(b^2-2b+2) \geq 4$ Solution:
$a,b>1, a -1= \frac{1}{b-1},a^2 - 2a +2 =(a-1)^2+1= \frac{1}{(b-1)^2}+1$ $\Longrightarrow (a^2 - 2a +2)(b^2 -2 b + 2) \geq 4 \iff\left(\frac{1}{(b-1)^2}+1\right)((b-1)^2+1)\geq 4$ $\iff (b-1)^2+ \frac{1}{(b-1)^2}\geq 2$

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#9 h Mar 10, 2025, 1:23 PM

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Let $a, b\geq 0$ and $a+2b+ab\geq \frac{17}{4} .$ Prove that
$a+2b \geq 5\sqrt 2-4$ $2a+3b \geq 5\sqrt 6-7$ $3a+4b \geq 10(\sqrt 3-1)$ Let $a, b\geq 0$ and $a+2b+3ab\geq \frac{73}{12} .$ Prove that
$a+2b \geq 3\sqrt 2-\frac43$ $2a+3b \geq 3\sqrt 6-\frac73$ $3a+4b \geq 6\sqrt 3-\frac{10}3$

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DAVROS

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#10 h Mar 10, 2025, 3:19 PM

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sqing wrote:

Let $a, b\geq 0$ and $a+2b+ab\geq \frac{17}{4} .$ Prove that $2a+3b \geq 5\sqrt 6-7$

solution

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#11 h Mar 10, 2025, 3:55 PM

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sqing wrote:

Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that $(a^2-a+b+1)(b^2-b+a+1) \geq 25$

solution

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#12 h Mar 11, 2025, 2:11 AM

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Very very nice.Thank DAVROS.
Let $a,b,c\geq 1$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+8 .$ Prove that
$ab+bc +ca\leq 27$ Let $a,b,c\geq 2.$ Prove that
$(a+1)(b+1)(c +1)-3abc\leq 3$ Let $a,b,c> 0 .$ Prove that
$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3)^2\geq 4(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3)^2\geq 24+4 (\frac{b}{a}+\frac{c}{b}+\frac{a}{c})$ Let $a,b\geq 0 .$ Prove that
$a^4+b^4 +1\geq ab(a+b+1)$ $a^5+b^5 +1\geq ab(a^2+b^2+1)$ $a^7+b^7 +1\geq ab(a+b^3+a^3b)$ $a^8+b^8 +1\geq ab(a+b^4+a^4b)$

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#13 h Mar 11, 2025, 6:22 AM

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sqing wrote:

Let $a, b\geq 0$ and $a+2b+3ab\geq \frac{73}{12} .$ Prove that $2a+3b \geq 3\sqrt 6-\frac73$

solution

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#14 h Mar 11, 2025, 7:05 AM

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Very very nice.Thank DAVROS.
Let $a,b,c\geq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+c}+\frac{1}{ac+b} \geq1$ $\frac{1}{ab+c+2}+\frac{1}{ac+b+2} \geq \frac{1}{2}$ Let $a,b,c> 0 .$ Prove that
$\frac{a}{2a+b+1}+ \frac{b}{2b+c+1}+ \frac{c}{2c+a+1}+ \frac{1}{a+b+c+1} \leq 1$ Let $a,b,c\geq 2 .$ Prove that
$(a^2+a+1)(b^2+b+1)(c^2+c+1)-5a^2b^2c^2\leq 23$ Let $a,b,c> 1$ and $a+b+c\leq 12 .$ Prove that
$\frac{a}{a^2-1}+\frac{b}{b^2-1}+\frac{c}{c^2-1}\geq \frac{12}{15}$ Let $a,b,c> 0$ and $a+b+c=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{25}{48abc+1}$ $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{81}{54abc+1}$ Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{16}{12ab+1}$ $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{64}{28ab+1}$

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#15 h Mar 11, 2025, 7:38 AM

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Let $a,b>0.$ Prove that
$ab (a^2+4b^2)\leq \frac{(41+22\sqrt[3] 2+24\sqrt[3]4)(a+6b)^4}{6000}$ $ab (a^2+4b^2)\leq \frac{(7129+1467\sqrt[3] 3+2241\sqrt[3]9)(a+b)^4}{3200}$ Let $a,b,c>0$ and $a^2=b^2+c^2.$ Prove that
$abc(6a^3+b^3+c^3)\leq \left(262-\frac{741}{2\sqrt2}\right)(a+b+c)^6$

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#16 h Mar 12, 2025, 12:27 PM

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

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#17 h Mar 12, 2025, 12:47 PM

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Let $a,b>0$ and $(a-3b+1)(b-3a+1)\geq 9.$ Prove that
$\frac{1}{a^2}+ \frac{2}{ab} +\frac{1}{b^2} \leq 1$

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#18 h Mar 13, 2025, 7:26 AM

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sqing wrote:

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

solution

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#19 h Mar 13, 2025, 8:19 AM

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sqing wrote:

Let $a,b>0$ and $(a-3b+1)(b-3a+1)\geq 9.$ Prove that $\frac{1}{a^2}+ \frac{2}{ab} +\frac{1}{b^2} \leq 1$

solution

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#20 h Mar 13, 2025, 12:14 PM

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Very very nice.Thank DAVROS.

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#21 h Mar 18, 2025, 3:13 PM

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Let $a,b,c\geq 0$ and $a^2+b^2 +c^2 =3.$ Prove that $\sqrt 6 - \frac{5}{2}\leq (a-1)(b-1)(c-1) \leq \sqrt 3 -1$

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#22 h Mar 19, 2025, 4:36 AM

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Let $a,b$ be reals such that $a^2+b^2 =4.$ Prove that
$\sqrt {5-2a}+ \sqrt {13-6b} \geq \sqrt {10}$ $3\sqrt {5-2a}+\sqrt {13-6b}\geq 2\sqrt {10}$

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#23 h Mar 21, 2025, 6:09 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a^2+b^2 +c^2 =3.$ Prove that $\sqrt 6 - \frac{5}{2}\leq (a-1)(b-1)(c-1) \leq \sqrt 3 -1$

solution

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#24 h Mar 21, 2025, 7:56 AM

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Very very nice.Thank DAVROS.

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#25 h Mar 21, 2025, 9:33 AM

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Let $a,b,c>0$ and $a^2+b^2+c^2+3\leq 2(ab+bc+ca).$ Prove that
$a+b+c\leq 3abc$ Let $a,b,c>0$ and $a^2+b^2+c^2+1\leq \frac{4}{3}(ab+bc+ca).$ Prove that
$a+b+c\leq 3abc$

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#26 h Mar 21, 2025, 10:08 AM

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sqing wrote:

Let $a,b,c>0$ and $a^2+b^2+c^2+1\leq \frac{4}{3}(ab+bc+ca).$ Prove that $a+b+c\leq 3abc$