But we can't compare them or can we?

$\dfrac {1}{\sqrt {1000}+\sqrt {1001}}<\dfrac {1}{\sqrt {1000}+\sqrt{999}}$

Rationalize the denominators to obtain

$\sqrt {1001}-\sqrt {1000}<\sqrt {1000}-\sqrt {999}$

Rearranging both sides yields

$\sqrt {1001}+\sqrt {999}<2\sqrt {1000}$

$f(x)=\sqrt{x}$ is (strictly) concave since $f'(x)=\frac{1}{2\sqrt{x}}$ is decreasing. Now $\sqrt{1000}>\frac{\sqrt{999}+\sqrt{1001}}{2}$ by definition of concavity.

$\sqrt{1001}+\sqrt{999}=\sqrt{1001+999+2\sqrt{1001*999}}=\sqrt{2000+2\sqrt{1000^2-1}}$ $2\sqrt{1000}=\sqrt{4000}$ $\sqrt{2000+2\sqrt{1000^2-1}}\boxed{?}\sqrt{4000}$ $2000+2\sqrt{1000^2-1}\boxed{?} 4000$ $\sqrt{1000^2-1}\boxed{?} 1000$ $1000^2-1\boxed{?} 1000^2$ $1000^2-1\boxed{<}1000^2$

We know that $\displaystyle \frac{x+y}{2} \leq \sqrt{\frac{x^2+y^2}{2}}$ (the QM-AM inequality). If we have $x = \sqrt{1001}$ and $y = \sqrt{999}$ , we get that

$\frac{\sqrt{1001}+\sqrt{999}}{2} \leq \sqrt{\frac{(\sqrt{1001})^2+(\sqrt{999})^2}{2}} = \sqrt{\frac{1001+999}{2}} = \sqrt{1000}$

Multiplying two on both sides we have

$\sqrt{1001}+\sqrt{999} \leq 2\sqrt{1000}$

Since both sides are not equal since the QM-AM inequality becomes an equality only when $x = y$ , we can conclude

$\boxed{\sqrt{1001}+\sqrt{999} < 2\sqrt{1000}}$

$\sqrt{x+1} - \sqrt{x}$ is a decreasing function. So $\sqrt{1001} - \sqrt{1000} < \sqrt{1000} - \sqrt{999} \implies \sqrt{1001} + \sqrt{999} \lt 2\sqrt{1000}$

√1001+√999=√(1000+1)+√(1000-1) by squaring the expression =1000+1+1000-1+2×√(1000+1)×√(1000-1)=2000+2×(√1000^2-1)=2000+2×√999999. {2×√1000}^2=4000=2000+2×√1000000 So {2×√1000}^2>{√1000+√999}^2. 2√1000>√1001+√999 two expressions are positive values .

I don't know if my logic is right, but:

$(\sqrt{1001} + \sqrt{999})^2$ and $(2\sqrt{1000})^2$

equals

$1001 + 999 < 4 \times 1000$

sqrt(2)=1.41 ; sqrt(3)=1.73 ; sqrt(4)=2

sqrt(2) + sqrt(4) = 3.41 < 3.46 = 2*sqrt(3)

If it worked for 2 and 4 it must work for 999 and 1001 :)

As $\sqrt{999}+\sqrt{1001}$ and $2\sqrt{1000}$ are positive numbers its relation won't change if we square them. $\sqrt{999}+\sqrt{1001} ¿ 2\sqrt{1000}$ $2000+2\sqrt{100089} ¿ 4000$ $\sqrt{400356} ¿ \sqrt{4000000}$ $400356 ¿ 4000000$ $400356 < 4000000$ This means the initial sign was $<$ , so $2\sqrt{1000}$ is greater

The rate of change of value of $\sqrt{n}$ , as n increases incrementally, decreases. This means that $\sqrt{1000}$ is going to be slightly above the average of $\sqrt{999}$ and $\sqrt{1001}$ , such that $2\sqrt{1000}$ is larger.

O (n) > O (n^0.5)

A change in the positive direction has less and less of an effect on a less than linear equation, so the sqrt(999) will be further from sqrt(1000) than sqrt(1001).

We can rewrite $\sqrt{1001} +\sqrt{999}$ as $(\sqrt{1000} + d_1) + (\sqrt{1000} - d_2)$ , and

$(\sqrt{1000} + d_1) + (\sqrt{1000} - d_2) = 2\sqrt{1000} + (d_1 - d_2)$

Now, since the difference between $\sqrt{n}$ and $\sqrt{n+1}$ becomes smaller as the value of $n$ increases, we have that $d_1 < d_2 \iff d_1 - d_2 < 0$

and so $2\sqrt{1000} + (d_1 - d_2) < 2\sqrt{1000} \implies \sqrt{1001} +\sqrt{999} < 2\sqrt{1000}$