Interesting inequalities

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sqing

#1 h Apr 27, 2025, 3:12 AM

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Let $a,b\geq 0$ and $a+b+ab=3.$ Prove that
$ab^2( b +1) \leq 4$ $ab( b +1) \leq \frac{9}{4}$ $a^2b ( a+b^2 ) \leq \frac{76}{27}$ $a^2b( b +1 ) \leq \frac{3(69-11\sqrt{33})}{8}$ $a^2b^2( b +1 ) \leq \frac{2(73\sqrt{73}-595)}{27}$

This post has been edited 1 time. Last edited by sqing, Apr 27, 2025, 4:53 AM

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sqing

#2 h Apr 27, 2025, 3:57 AM

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Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^2b^2( ka^2+b^2)(a^2+kb^2) \leq(k+1)^2$ Where $k>0.$

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#3 h Apr 27, 2025, 6:48 AM

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sqing wrote:

Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^2b^2( ka^2+b^2)(a^2+kb^2) \leq(k+1)^2$ Where $k>0.$

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sqing

#4 h Apr 27, 2025, 7:50 AM

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Good.Thanks.

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sqing

#5 h Apr 27, 2025, 12:34 PM

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Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^2b^2( 2a^2+ab+b^2)(a^2+ab+2b^2)\leq 16$ $a^2b^2( 2a^2+a+b^2)(a^2+b+2b^2) \leq 16$ $a^2b^2(2a^2+3b^2)(3a^2+2b^2) \leq 25$ $a^2b^2( 2a^2+a+3b^2)(3a^2+b+2b^2) \leq 36$

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#6 h Apr 28, 2025, 2:17 AM

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sqing wrote:

Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^2b^2( 2a^2+ab+b^2)(a^2+ab+2b^2)\leq 16$

Using the AM-GM inequality, we have:
$a^2b^2(2a^2+ab+b^2)(a^2+ab+2b^2) \leq (ab)^2 \cdot \frac{(3a^2+2ab+3b^2)^2}{4}$ $= \frac{[4 \cdot 8ab \cdot (3a^2+2ab+3b^2)]^2}{4096} \leq \frac{(3a^2+10ab+3b^2)^4}{4096}$ $\leq \frac{(4a^2+8ab+4b^2)^4}{4096} = \frac{256 \cdot 256}{4096} = 16.$ We have equality if and only if $a = b = 1$ .

This post has been edited 3 times. Last edited by pooh123, Apr 28, 2025, 5:58 AM
Reason: typo

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#7 h Apr 28, 2025, 2:22 AM

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sqing wrote:

Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^2b^2(2a^2+3b^2)(3a^2+2b^2) \leq 25$

Using the AM-GM inequality, we have:
$a^2b^2(3a^2+2b^2)(2a^2+3b^2) \leq (ab)^2 \cdot \frac{(5a^2+5b^2)^2}{4}$ $= \frac{25}{16} \cdot [2ab(a^2+b^2)]^2 \leq \frac{25}{16} \cdot \frac{(a+b)^8}{16} = \frac{25}{16} \cdot \frac{256}{16} = 25.$ We have equality if and only if $a = b = 1$ .

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sqing

#8 h Apr 28, 2025, 8:28 AM

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Very nice.Thanks.

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#9 h Apr 28, 2025, 3:03 PM

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sqing wrote:

Let $a,b\geq 0$ and $a+b+ab=3.$ Prove that
$ab^2( b +1) \leq 4$

Since $(a+1)(b+1)=4$ , then by AM-GM: $\frac{9(a+1)+4(b+1)}{2} \ge \sqrt{9\cdot 4\cdot 4} = 24 \implies 9a+4b+1 \ge 12 = 4(a+b+ab).$ Thus $5a+1 \ge 4ab \implies a+1 \ge 4a(b-1) \implies 4 \ge 4a(b^2-1) \implies ab^2 \le a+1 \implies ab^2(b+1) \le 4,$ as desired.

This post has been edited 1 time. Last edited by aaravdodhia, Apr 28, 2025, 3:05 PM

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sqing

#10 h Apr 28, 2025, 3:08 PM

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Very very nice.Thanks.

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