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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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6  
@Shawn: For all $x$? Surely not, put $x=a$. The assertion becomes $\sqrt{2a} >a$, which is false if $a \ge 2$. Can you reformulate the question? Maybe you are trying to find what values of $x$ satisfy the inequality? – André Nicolas May 5 '11 at 3:20
    
Oops sorry the question is to solve for such for x to make this true. – Shawn May 5 '11 at 3:57
up vote 0 down vote accepted

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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Ah thanks for the solution, I guess I forgot you could simplify roots by multiplying their inner values. – Shawn May 5 '11 at 5:01
1  
Let $a=135/128$ then your equation gives $x>-\frac{405 \sqrt{5655}}{32768}\approx -0.9294$ but $x=-1$ works... – Listing May 5 '11 at 5:02
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The third step assumes that $a-2\gt 0$ when it passes from $2\sqrt{a^2-x^2}\gt a(a-2)$ to $4(a^2-x^2)\gt a^2)(a-2)^2$. – Arturo Magidin May 5 '11 at 5:26
    
@user3123, @Arturo : oops.. – JiminP May 5 '11 at 5:57

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 $0<a<2$, 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. – Listing May 5 '11 at 5:11

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