Prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$

Question

How can I prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$?

I can't convert the right side to get the $\sqrt 5$.

Answer 1

Hint:

$$(\sqrt{7}+2\sqrt{5})^2-(2+\sqrt{35})^2=?$$

Answer 2

If $\sqrt7=a,\sqrt 5=b$

We need $$a+2b<2+ab$$

$$\iff0>a-2-b(a-2)=(a-2)(a-b)$$

But $a=\sqrt7>\sqrt5(=b)>2$

Answer 3

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

$$\sqrt7+2\sqrt5\lt{11\over4}+2\cdot{9\over4}={29\over4}\lt{30\over4}={15\over2}=2+{11\over2}\lt2+\sqrt{35}$$

Answer 4

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 $-\sqrt{5}<-1$.

Answer 5

Since $\sqrt5\gt1$ and $\sqrt7\gt2$, we have

$$\sqrt7-2\lt\sqrt5(\sqrt7-2)=\sqrt{35}-2\sqrt5$$

and therefore $\sqrt7+2\sqrt5\lt2+\sqrt{35}$.

Answer 6

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?)

$√7 +2√5 < 2 +\sqrt{35}.$

Answer 7

Why make things more complex than they are? As both sides are positive, just compare the squares: $$ \sqrt 7+2\sqrt 5<2+\sqrt{35}\iff7+4\sqrt{35}+4\cdot5<4+4\sqrt35+35 \iff 27 <39. $$