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sqing

#1 h Yesterday at 3:45 AM

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Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$ Where $4\leq k\in N^+.$
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ Where $16\geq k>0 .$

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#2 h Yesterday at 3:59 AM

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sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$ Where $5\leq k\in N^+.$

$\frac{1}{a}+\frac{1}{b}-\frac{2k}{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}}=\frac{a+b}{ab}-\frac{2k}{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}}=\frac{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}-2kab}{ab\left( 1+{{k}^{2}}{{a}^{2}}{{b}^{2}} \right)}=\frac{{{\left( kab-1 \right)}^{2}}}{ab\left( 1+{{k}^{2}}{{a}^{2}}{{b}^{2}} \right)}\ge 0,\forall k\in \mathbb{R}$

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sqing

#3 h Yesterday at 1:14 PM

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Nice.Thanks.

This post has been edited 1 time. Last edited by sqing, Yesterday at 1:21 PM

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#4 h Yesterday at 1:19 PM

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Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ Where $16\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$

This post has been edited 3 times. Last edited by sqing, Yesterday at 1:53 PM

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#5 h Yesterday at 1:29 PM

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sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$

$a+b=1$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ <===> $a^2b^2- \frac{17}{4}ab+1 \geq 0$ <===> $ab \leq \frac{1}{4}$

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sqing

#6 h Yesterday at 1:48 PM

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Nice.Thanks.

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#8 h Yesterday at 1:54 PM

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sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ Where $16\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$

$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ $(4+ \frac{k}{4})ab \leq 1+ka^2b^2$ <===> $(kab-4)(4ab-1) \geq 0$ <===> $ab \leq \frac{1}{4}$

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sqing

#9 h Yesterday at 2:46 PM

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Nice.Thanks.

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