# 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 ? Sep 10, 2014 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. Sep 10, 2014 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 :-) Sep 10, 2014 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. Sep 11, 2014 at 11:35
• @CarlWitthoft. Thanks ! I learnt a lot from your answers and comments. Cheers :-) Sep 11, 2014 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 $2x + 7 + 2\sqrt{x^2 +7*x}$ Sep 10, 2014 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 $[0, ({29 \over 12})^2]$. Sep 10, 2014 at 17:36