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sqing

#1 h Today at 2:49 AM

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Let $a,b > 0$ and $ab\geq 1 .$ Prove that
$\frac{a^2+kab+b^2}{ab+1}\ge \frac{k+2}{2}\sqrt{2-\frac{1}{ab}}$ Where $-2< k\leq 2 .$
$\frac{a^2-ab+b^2}{ab+1}\ge \frac{1}{2}\sqrt{2-\frac{1}{ab}}$ $\frac{a^2+ab+b^2}{ab+1}\ge \frac{3}{2}\sqrt{2-\frac{1}{ab}}$ $\frac{(a+b)^2}{ab+1}\ge 2\sqrt{2-\frac{1}{ab}}$

This post has been edited 1 time. Last edited by sqing, Today at 2:56 AM

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#2 h Today at 4:48 AM

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sqing wrote:

Let $a,b > 0$ and $ab\geq 1 .$ Prove that
$\frac{a^2+kab+b^2}{ab+1}\ge \frac{k+2}{2}\sqrt{2-\frac{1}{ab}}$ Where $-2< k\leq 2 .$
$\frac{a^2-ab+b^2}{ab+1}\ge \frac{1}{2}\sqrt{2-\frac{1}{ab}}$ $\frac{a^2+ab+b^2}{ab+1}\ge \frac{3}{2}\sqrt{2-\frac{1}{ab}}$ $\frac{(a+b)^2}{ab+1}\ge 2\sqrt{2-\frac{1}{ab}}$

According to the AM-GM inequality,
$2ab = \frac{2(1+ab) + (4ab-2)}{3} \ge \sqrt[3]{(1+ab)^2 (4ab-2)},$ hence
$\frac{k+2}{2} \sqrt{2-\frac{1}{ab}} = \frac{k+2}{2} \sqrt{\frac{4ab-2}{2ab}} \le \frac{k+2}{2} \cdot \frac{2ab}{1+ab} = \frac{(k+2)ab}{1+ab} \le \frac{a^2+kab+b^2}{ab+1}$ holds for $ab\ge\frac{1}{2} , k \ge -2$ .

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sqing

#3 h Today at 6:52 AM

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Very nice.Thank lbh_qys:

Let $a,b > 0$ and $ab\geq \frac{1}{2} .$ Prove that
$\frac{a^2+kab+b^2}{ab+1}\ge \frac{k+2}{2}\sqrt{2-\frac{1}{ab}}$ Where $k\geq - 2 .$

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