Show that $\sqrt{10}+\sqrt 7>\sqrt{17}$

Question

I'm not actually sure what is there to show. It is obvious and can be verified with a calculator.

Is there a formal way of proving/showing this?

Answer 1

Since:

$$10+7+2\cdot\sqrt{10} \cdot \sqrt{7} > 17$$ $$(\sqrt{10} + \sqrt{7})^2 > (\sqrt{17})^2$$

and since both sides are positive:

$$\sqrt{10} + \sqrt{7} > \sqrt{17}$$

Answer 2

By AM-GM inequality

$$\frac{\sqrt{10}+\sqrt 7}{2}\ge \sqrt{\sqrt{70}}\iff \sqrt{10}+\sqrt 7\ge 2\sqrt{\sqrt{70}}$$

and

$$2\sqrt{\sqrt{70}}=2\sqrt[4]{70}>2\sqrt[4]{64}= 2\sqrt 8=\sqrt{32}>\sqrt{17}$$

Answer 3

You can begin by squaring:

$$(\sqrt{10}+\sqrt7)^2=10+2\sqrt{10}\cdot\sqrt7+7=17+2\sqrt{70}$$ and now just observe that the square root of any positive number is poitive itself...

Answer 4

First notice $0 < a < b$ then $a^2 = a*a < a*b < b*b < b^2$ and if $0 < a, 0 < b$ but $a \ge b$ then $a^2 = a*a \ge a*b \ge b*b \ge b^2$ so

Lemma: for $a >0; b> 0$ then $a < b \iff a^2 < b^2$.

So ... square both sides. $(\sqrt{10} + \sqrt{7})^2 = \sqrt{10}^2 + 2\sqrt{10}\sqrt{7} + \sqrt{7}^2 = 10 + 2\sqrt{10}\sqrt{7} +7 =17 + 2\sqrt{10}\sqrt{7} > 17 = \sqrt{17}^2$.

So $\sqrt{10} + \sqrt 7 > \sqrt{17}$.

It is obvious

It's not obvious to me: $3 < \sqrt{10} < 4$ and $2 < \sqrt{7} < 3$ so $5< \sqrt{10} + \sqrt{7} < 7$ and $4 < \sqrt{17} < 5$ and .... oh, I guess it is obvious .... $4 < \sqrt{17} < 5 = 3 + 2 < \sqrt{10} + \sqrt{7} < 7$

...and can be verified with a calculator.

Calculators are magic and we aren't allowed to use magic until we have proven magic works and is trustworthy. So unless you can tell me why when I press in $\sqrt {10} + \sqrt{7}$ I get $3.1622776601683793319988935444327 + 2.6457513110645905905016157536393$ and when I press in $\sqrt 7 =4.1231056256176605498214098559741$, I won't trust it and won't let you cite that as a proof.

However if you can tell my why the calculators do that I can accept it as a proof.

I mean.... if we were allowed to say "My calculator told me so" we ought to be able to say "A really smart grown-up told me so" and ... well, there's no point in going to school at all then.

Answer 5

As an alternative

$$\sqrt{10}+\sqrt 7>\sqrt{9}+\sqrt 4=3+2=5=\sqrt{25}>\sqrt{17}$$

Answer 6

$f(x)=\sqrt{x}$ is a concave function and $(17,0)\succ(10,7)$.

Thus, by Karamata $$\sqrt{10}+\sqrt{7}>\sqrt{17}+\sqrt{0}=\sqrt{17}.$$