Prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$ How can I prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$?
I can't convert the right side to get the $\sqrt 5$.

Since $2^2 <7$. we get 
$2=\sqrt{2^2} <√7$ since $√$ is strictly increasing.
Also since $1 < 5$ we get $1 <√5.$
Finally :
$2(√5-1) < √7(√5-1)$; (Why?)
$√7 +2√5 < 2 +\sqrt{35}.$
