# Problem Of The Week #448 December 28th 2020

Status
Not open for further replies.

Staff member

#### anemone

##### MHB POTW Director
Staff member
No one answered last week's POTW. However, I will give the community another week's time to take another stab at the problem. I am looking forward to receiving submissions from the members!

• topsquark

#### anemone

##### MHB POTW Director
Staff member
No one answered last two week's POTW. However, you can find the suggested solution (from other) as follows:

The solution set consists of all non-negative real numbers, as we shall show in the following.

Note that we need $x\ge 0$ in order for the right hand side of the inequality to be defined. Moreover, for all non-negative real numbers $x$, we have

\begin{align*}\sqrt{4x^2-8x+5}+\sqrt{3x^2+12x+16}&=\sqrt{4(x-1)^2+1}+\sqrt{3(x+2)^2+4}\\& \ge 1+2 \\&=3\end{align*}

On the other hand, $6\sqrt{x}-x-6=-(\sqrt{x}-3)^2+3\le 3$. This completes the proof.

• topsquark
Status
Not open for further replies.