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sqing

#1 h Mar 22, 2025, 2:00 PM

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Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{16}}{1+ ka^3b^3}$ Where $32\geq k>0 .$
Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{64}}{1+ ka^4b^4}$ Where $\frac{256}{3}\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+16a^3b^3}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{6}{1+32a^3b^3}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+64a^4b^4}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{16}{3}}{1+\frac{256}{3}a^4b^4}$

This post has been edited 2 times. Last edited by sqing, Mar 22, 2025, 2:37 PM

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#2 h Mar 22, 2025, 2:22 PM

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sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{16}}{1+ ka^3b^3}$ Where $32\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+16a^3b^3}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{6}{1+32a^3b^3}$

put $x=ab$ $(x-\frac{1}{4})(16kx^2+4kx-64) \geq 0$ <===> $(x-\frac{1}{4})(4kx^2+kx-16) \geq 0$ because: $32\geq k>0 .$
so: $x_1x_2=-\frac{4}{k}$ $x_1+x_2=\frac{1}{4}$ <===> $ab \leq \frac{1}{4}$ done.

This post has been edited 1 time. Last edited by SunnyEvan, Mar 22, 2025, 2:23 PM

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sqing

#3 h Mar 22, 2025, 2:24 PM

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Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{256}}{1+ ka^5b^5}$ Where $256\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+256a^5b^5}$ Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{1024}}{1+ ka^6b^6}$ Where $\frac{4096}{5}\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{24}{5}}{1+\frac{4096}{5}a^6b^6}$

This post has been edited 2 times. Last edited by sqing, Mar 22, 2025, 3:03 PM

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#5 h Mar 22, 2025, 2:54 PM

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Let $a,b> 0$ and $a+b=1 .$ Prove that :
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4^n}}{1+ ka^{n+1} b^{n+1}}$ Where $0<k \leq \frac {4^{n+1}}{n}$ $n \in N^*$

This post has been edited 2 times. Last edited by SunnyEvan, Mar 22, 2025, 3:05 PM

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sqing

#6 h Mar 22, 2025, 3:09 PM

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Yeah.Thanks.
Let $a,b> 0$ and $a+b=1 .$ Prove that :
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4^n}}{1+ ka^{n+1} b^{n+1}}$ Where $n \in N^+$ and $0<k \leq \frac {4^{n+1}}{n} .$

This post has been edited 1 time. Last edited by sqing, Mar 22, 2025, 3:12 PM

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#7 h Mar 22, 2025, 3:12 PM

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That is so interesting ! :-D

:-D

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sqing

#8 h Mar 22, 2025, 3:19 PM • 1 Y

Y by SunnyEvan

sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that $\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4096}}{1+ ka^7b^7}$ Where $\frac{8192}{3}\geq k>0 .$

Solution of DAVROS:
Equivalent to $\frac1x+kx^6 \ge 4+\frac{k}{4096}$ where $x=ab\le\frac14$

$\text{LHS} =6\cdot \frac{k}{4^7x}+kx^6+ \left(1-\frac{6k}{4^7}\right)\frac1x \ge 7\sqrt[7]{\frac{k^7}{4^{42}}}+ \left(1-\frac{6k}{4^7}\right)\frac1x$

$\text{LHS} \ge \frac{7k}{4^6} + 4\left(1-\frac{6k}{4^7}\right)=4+\frac{k}{4096}$ provided $k \le \frac{4^7}6 =\frac{8192}3$

Let us continue to work hard and explore further.

This post has been edited 1 time. Last edited by sqing, Mar 22, 2025, 4:15 PM

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#9 h Mar 22, 2025, 3:24 PM

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OK , let's go! :-D

:-D

This post has been edited 1 time. Last edited by SunnyEvan, Mar 22, 2025, 3:25 PM

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