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prove $(a + x)^{1/2} + (a - x)^{1/2} \gt a$ for any real $a\gt 0$.

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    Oops sorry the question is to solve for such for$x$to make this true.2011-05-05

2 Answers 2

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Obviously $-a\lt x\lt a$ (Comment of Shawn).

Squaring both sides... $(a+x)+(a-x)+2\sqrt{a^2 - x^2} = 2a + 2\sqrt{a^2-x^2} > a^2$

or $2\sqrt{a^2-x^2} > a(a-2)$

(When $a\geq2$)

$4(a^2-x^2) > a^2 (a^2 - 4a + 4)$

or $a^4 - 4a^3 + 4x^2 < 0$

or $x^2 < a^3 - a^4/4$

or $-a \sqrt{a - a^2/4} < x < a \sqrt{a - a^2/4}$.

(When $a\leq2$) (Comment of Arturo Magidin)

Since $a(a-2)\lt0$, $-a\lt x \lt a$ is enough.

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    @user3123, @Arturo : oops..2011-05-05
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Squaring both sides, we obtain the equivalent inequality $2\sqrt{a^2-x^2} >a^2-2a$ If $a^2-2a<0$, that is, if $0, the inequality automatically holds for all $x$ at which the left-hand side is defined, that is, for all $x$ such that $|x| \le a$.

So now consider the case $a\ge 2$. Then the inequality $2\sqrt{a^2-x^2}>a^2-2a$ is equivalent to $4(a^2-x^2)>(a^2-2a)^2$. This simplifies to $4x^2 <4a^3-a^4$ or equivalently $|x|<(a/2)\sqrt{4a-a^2}$. Note that in particular there are no solutions if $a \ge 4$.

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    This one is the correct answer.2011-05-05