# Inequality involving square roots

I need help with this inequality:

$\sqrt x +\sqrt{x+7} + 2\sqrt{x^2+7x} <35-2x$

It doesn't seem solvable. All roots of the corresponding equation are irrational.

• What about $0 \leq x < \frac{841}{144}$ which the only root for the equation ? – Claude Leibovici Sep 10 '14 at 12:22
• @ClaudeLeibovici The existence of a single (real) root doesn't guarantee that f(x) < 35 everywhere else. In the general case you need to show the function is positive for larger x , I think. – Carl Witthoft Sep 10 '14 at 17:17
• @CarlWitthoft.You are perfectly correct, indeed. I was just mentionning that there is only one root which is rational. I wonder how this rationality could be established. Cheers :-) – Claude Leibovici Sep 10 '14 at 20:05
• @ClaudeLeibovici well, once you've manipulated the inequality into an equivalent polynomial expression, there's some theorem about the number of real roots based on the number of coefficient sign changes. – Carl Witthoft Sep 11 '14 at 11:35
• @CarlWitthoft. Thanks ! I learnt a lot from your answers and comments. Cheers :-) – Claude Leibovici Sep 11 '14 at 11:46

Let $a=\sqrt{x}+\sqrt{x+7}$, then $a+a^2<42$, and $-7<a<6$
• @DeanMacGregor $a^2 = x + 2\sqrt{x}*\sqrt{x+7} + (x+7)$ which reduces to $2x + 7 + 2\sqrt{x^2 +7*x}$ – Carl Witthoft Sep 10 '14 at 17:15
• Note $\sqrt{x} + \sqrt{x + 7}$ is monotone in $x$, and that it is only defined for $x \geq 0$. So you need to solve $\sqrt{x} + \sqrt{x + 7} = 6$ and then the solution interval will be $[0,x]$. If you work it out, $x = ({29 \over 12})^2$, so the final answer is $[0, ({29 \over 12})^2]$. – Zarrax Sep 10 '14 at 17:36