Problem Of The Week #448 December 28th 2020

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anemone

MHB POTW Director
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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!

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.

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