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 $\sqrt 5$.
 A: Why make things more complex than they are? As both sides are positive, just compare the squares:
$$ \sqrt 7+2\sqrt 5<2+\sqrt{35}\iff7+4\sqrt{35}+4\cdot5<4+4\sqrt35+35 
\iff 27 <39. $$
A: Hint:
$$(\sqrt{7}+2\sqrt{5})^2-(2+\sqrt{35})^2=?$$
A: If $\sqrt7=a,\sqrt 5=b$
We need 
$$a+2b<2+ab$$
$$\iff0>a-2-b(a-2)=(a-2)(a-b)$$
But $a=\sqrt7>\sqrt5(=b)>2$
A: Just to give a different approach, from $7\cdot16=112\lt121=11^2$ we have $\sqrt7\lt11/4$, from $5\cdot16=80\lt81=9^2$ we have $\sqrt5\lt9/4$, and from $11^2=121\lt140=4\cdot35$ we have $11/2\lt\sqrt{35}$.  It follows that
$$\sqrt7+2\sqrt5\lt{11\over4}+2\cdot{9\over4}={29\over4}\lt{30\over4}={15\over2}=2+{11\over2}\lt2+\sqrt{35}$$
A: Start from the following inequality (I obtained it from rearranging the intial inequality as you can see in the following steps)
$$(\sqrt{7}-2)(1-\sqrt{5})<0$$
$$\implies \sqrt{7}-2-\sqrt{7}\sqrt{5}+2\sqrt{5}<0$$
$$\implies \sqrt{7}+2\sqrt{5}<2+\sqrt{35}.$$
The first inequality is trivally true as $\sqrt{7}>\sqrt{4}=2$ and $-\sqrt{5}<-1$.
A: Since $\sqrt5\gt1$ and $\sqrt7\gt2$, we have
$$\sqrt7-2\lt\sqrt5(\sqrt7-2)=\sqrt{35}-2\sqrt5$$
and therefore $\sqrt7+2\sqrt5\lt2+\sqrt{35}$.
A: 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?)
$√7 +2√5 < 2 +\sqrt{35}.$
