Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$

Solution 1:

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

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

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

Solution 2:

This is an overkill solution, but this is an immediate application of Cauchy-Schwarz:

$$\left( 1 \cdot \sqrt{1001}+1 \cdot\sqrt{999}\right)^2 \leq (1+1)(1001+999)=4000\,.$$

Solution 3:

The answer is: YES, we can!

$$ \begin{align} (\sqrt{1001}+\sqrt{999})^2&=2000+2\sqrt{1001\times 999} \\ &=2000+2\sqrt{(1000+1)(1000-1)} \\ &=2000+2\sqrt{1000^2-1} \end{align} $$ \begin{align} \text{and that: }(2\sqrt{1000})^2&=4000 \\ &=2000+2\sqrt{1000^2} \end{align}

Since $2000+2\sqrt{1000^2-1}<2000+2\sqrt{1000^2}$

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

Solution 4:

You can tell without calculation if you can visualize the graph of the square-root function; specifically, you need to know that the graph is concave (i.e., it opens downward). Imagine the part of the graph of $y=\sqrt x$ where $x$ ranges from $999$ to $1001$. $\sqrt{1000}$ is the $y$-coordinate of the point on the graph directly above the midpoint, $1000$, of that interval. $\frac12(\sqrt{999}+\sqrt{1001})$ is the average of the $y$-coordinates at the ends of this segment of the graph, so it's the $y$-coordinate of the point directly above $x=1000$ on the chord of the graph joining those two ends. The concavity of the graph shows that the chord lies below the graph. So $\frac12(\sqrt{999}+\sqrt{1001})<\sqrt{1000}$. Multiply by $2$ to get the numbers in your question.