Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
@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.

share|cite|improve this answer
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
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
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 $a \ge 4$.

share|cite|improve this answer
This one is the correct answer. – Listing May 5 '11 at 5:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.