<s>[INST] How $10 +2\sqrt{2} \sqrt{10}+2 > 17$ becomes $2\sqrt{2} \sqrt{10} > 5$ due to theorem $3$? I'm reading Kazarinoff's Analytic inequalities. He gives the theorem 3 and 6:


*

*If $a>b$ and $c>d$ then $a+c> b+d$.

*If $a>b>0$, then $a^{\frac{p}{q}}> b^{\frac{p}{q}}$.
There is an exercise a little bit further: Show that $\sqrt{10}+\sqrt{2}> \sqrt{17}$.
And the solution goes as following: 
$$10 +2\sqrt{2} \sqrt{10}+2 > 17\tag{$\text{Theorem 6}$}$$
$$2\sqrt{2} \sqrt{10} > 5\tag{$\text{Theorem 3}$}$$
How is subtracting from both sides a consequence of theorem $3$ (Supposing that this is what is being done)? I took:
$$12 +2\sqrt{2} \sqrt{10} > 17$$
Then $12$ or $2\sqrt{2}\sqrt{10}$ might be $a$ or $c$ and writing $17$ as $12+5$, I'd have: 
$$12 +2\sqrt{2} \sqrt{10} > 12+5$$
If $a=12$, then by theorem $3$, $b=5$ to achieve $12>5$ according to the theorem, $c=2\sqrt{2} \sqrt{10}$, $d=12$ but to use the theorem, shouldn't $2\sqrt{2}\sqrt{10}>12$? Also, I could choose $a=2\sqrt{2}\sqrt{10}$ but the only reasonable choice for $c$ would be $5$ and I'd have to have $12>12$. 
I looked at Wikipedia and found the following:


  Which seems natural (and I am acquainted with its existence), but I'm a bit confused at how he deduced that from theorem $3$. 

I am aware that perhaps, he deduced something  that looks like a subtraction from both sides but is quite different, but I couldn't imagine more and all my experiments failed.
 [/INST] It does not follow directly from Theorem 3, but one can make a corollary:

$a>b$ $\implies$ $a+c>b+c$

Proof: By Definition 1, p. 2, $a>b$ means $a=b+h$, $h>0$. As $h/2>0$ too (by Theorem 4 that says $a>b$, $c>0$ $\implies$ $a/c>b/c$), the same definition gives that $a>b+h/2$. Now take $c$ and $d=c-h/2$. We have $c>d$, then by Theorem 3 
$$
a+c>b+h/2+d=b+c.
$$
</s>