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I've got stuck at these problem :

Prove that for any $x \geq 1$ and any $y \geq 1$ the following inequality is true: $$ {x}{\sqrt{y - 1}} + {y}{\sqrt{x - 1}} \leq xy. $$

The first thing that came into my mind was the AM-GM inequality (the extended version - I don't know if it has a specific name, other than mean inequality): $$ HM \leq GM \leq AM \leq SM $$ where $HM$, $GM$, $AM$, and $SM$ refer to the harmonic, geometric, arithmetic, and square mean, respectively.

I assume that's what I need to use, but I didn't get it right.

I would apreciate some hints.

Thanks!

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  • $\begingroup$ The extended version of AM-GM is called the power mean inequality i think. $\endgroup$ – Yiyuan Lee Oct 3 '15 at 9:08
  • $\begingroup$ You mean $x,y>1$. $\endgroup$ – user236182 Oct 3 '15 at 9:10
  • $\begingroup$ The inequality is false. Let $x=y=1.25$. $\endgroup$ – user236182 Oct 3 '15 at 9:13
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    $\begingroup$ Let $a^2 = x-1, b^2 = y-1$, then you want to show $$a+\frac1a+b+\frac1b \le (a^2+1)(b^2+1)$$ which will not hold for any $a, b \in (0, 1)$ as the LHS is clearly larger than $4$, while the RHS is lesser than $4$. $\endgroup$ – Macavity Oct 3 '15 at 9:35
  • $\begingroup$ @Macavity someone edited the post but changhed the inequality. There aren't any fractions line. I've just edited the post to the correct form. Please check it out. $\endgroup$ – scummy Oct 3 '15 at 10:28
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With the current version of the problem again let $a^2 = x-1, b^2 = y-1$, then you want to show $$(a^2+1)b + (b^2+1)a \le (a^2+1)(b^2+1)$$ $$\iff (b^2+1-b)a^2-(b^2+1) \cdot a+(b^2+1-b) \ge 0$$ $$\iff a^2-\frac{b^2+1}{b^2+1-b} \cdot a+1 \ge 0$$ $$\iff \left(a-\frac{b^2+1}{2(b^2-b+1)} \right)^2+\frac{(b-1)^2(3b^2-2b+3)}{4(b^2-b+1)^2} \ge 0$$ which is obvious as $3b^2-2b+3 > 0$. Equality is possible iff $a=b=1 \implies x=y=2$.

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  • $\begingroup$ I have a question : what if $x,y ∉ Q$? Then, can I still say that $a^2=x−1$ and $b^2=y−1$? What kind of numbers are $a,b$? $\endgroup$ – scummy Oct 5 '15 at 15:49
  • $\begingroup$ @scummy Never assumed $x,y$ are rational! this works for all Real numbers $x,y\ge 1$. Then $a,b$ are non-negative reals. $\endgroup$ – Macavity Oct 5 '15 at 15:53
  • $\begingroup$ Thanks, I've cleared up! Silly question. $\endgroup$ – scummy Oct 5 '15 at 15:57
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As Macavity correctly points out, the transformation $x=a^2+1, y=b^2+1$ reveals that the inequality does not hold for $x,y \in (1,2)$ (the LHS being $>4$, while the RHS being $<4$), and the same reasoning proves the inequality for $x,y \in (2,\infty)$.

For $x \in (1,2)$ and $y \in (2 ,\infty)$, the valitity depends on the concrete values of $x,y$.

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  • $\begingroup$ Someone edited the problem and changed something. There aren't any fractions. I've just edited the post (again). Please check it out. $\endgroup$ – scummy Oct 3 '15 at 10:24
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The given inequality is equivalent to:

$$ x,y\geq 1\Longrightarrow \frac{1}{x\sqrt{x-1}}+\frac{1}{y\sqrt{y-1}}\leq 1$$ but that cannot hold if $x$ or $y$ is close to one, since the function $f(x)=\frac{1}{x\sqrt{x-1}}$ is unbounded in a right neighbourhood of $1$.

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