## Monday, December 28, 2015

### Analysis Quiz 16: Multiple-Choice Math (Square Root Inequality)

Analysis Quiz 16: Multiple-Choice Math (Square Root Inequality)

Question 1: Which of the methods below do you think can be used to prove $\sqrt{3}-\sqrt{2}\gt \sqrt{4}-\sqrt{3}$?
A. AM-GM Inequality.
B. Jensen's Inequality.
C. Cauchy–Schwarz inequality.
D. Squaring both sides of the equation and squaring again to remove the square root to make numerical comparison.

Actually the given inequality can be proved correct through AM-GM inequality, Jensen inequality, Cauchy-Schwarz inequality and the most easy method, the squaring both sides of the equation (for as many time as needed to eliminate the square root) and lastly to compare the final integers at both sides of the inequality to see if the inequality sign agrees with the result.

In other words, all of them could be used to prove the target inequality and we should check all of the options above.

Question 2: If Jensen's inequality is the choice you made, what obstacles would you think you would encounter?
A. We would end up with proving $\sqrt{4}+\sqrt{2}\gt \sqrt{3}$ but this has not helped in proving $\sqrt{3}+\sqrt{3}\gt \sqrt{4}+\sqrt{2}$.
B. The function $f(x)=\sqrt{x}$ is not concave.
C. It's hard to prove the function $f(x)=\sqrt{x}$ is concave.
D. None of the above.

It's not hard to prove $f(x)=\sqrt{x}$ is concave (since  and therefore, $f\left(\dfrac{x_1+x_2}{2}\right)\gt \dfrac{f(x_1)+f(x_2)}{2}$. Putting in $x_1=2$ and $x_2=4$, we get

$f\left(\dfrac{2+4}{2}\right)\gt \dfrac{f(2)+f(4)}{2}$

$2f\left(3\right)\gt f(2)+f(4)$

$2\sqrt{3}\gt \sqrt{2}+\sqrt{4}$

$\sqrt{3}-\sqrt{2}\gt \sqrt{4}-\sqrt{3}$ and the proof is then follows.

Question 3: At first sight, AM-GM inequality doesn't seem helpful at all in proving the given inequality is true.
A. Yes.
B. No.

Yes, because if we're to use the AM-GM inequality based on what we're being asked, we most definitely would try out the following:

$\sqrt{2}+\sqrt{4}\gt 2\sqrt{\sqrt{2}\sqrt{4}}$

But we're required to prove

$2\sqrt{3}\gt \sqrt{2}+\sqrt{4}$

So we would reach an impasse if we proceed with what we have started.

Question 4: If you're given the hint to use AM-GM inequality as in proving $2+4\gt 2\sqrt{2}\sqrt{4}$, how would you proceed to prove the given inequality is true?
A. Squaring both sides of the inequality.
B. Add $2$ and $4$ to both sides of the inequality.
C. Subtract $2$ and then $4$ from both sides of the inequality.

First, we know we wanted to prove $2\sqrt{3}\gt \sqrt{2}+\sqrt{4}$, with the sum of $\sqrt{2}$ and $\sqrt{4}$ on the right side of the inequality sign, next, from the hint, we see that the right side has the product of $\sqrt{2}$ and $\sqrt{4}$, therefore, our effort should remain on turning the product to the sum:

$2+4\gt 2\sqrt{2}\sqrt{4}$

$2+2+4+4\gt 2+2\sqrt{2}\sqrt{4}+4$

$12\gt (\sqrt{2}+\sqrt{4})^2$

$4(3)\gt (\sqrt{2}+\sqrt{4})^2$

$2\sqrt{3}\gt \sqrt{2}+\sqrt{4}$

$\therefore \sqrt{3}-\sqrt{2}\gt \sqrt{4}-\sqrt{3}$ and the proof is then follows.

Question 5: Cauchy-Schwarz inequality is the most straightforward method to prove the given inequality to be true.
A. Yes.
B. No.

Yes, there is nothing to contemplate except for applying directly of the Cauchy-Schwarz inequality to $\sqrt{4}+\sqrt{3}$, and we get the result:

$\sqrt{4}+\sqrt{2}=\sqrt{2}\sqrt{2}+\sqrt{1}\sqrt{2}\lt \sqrt{2+1}\sqrt{2+2}=2\sqrt{3}$

$\therefore \sqrt{3}-\sqrt{2}\gt \sqrt{4}-\sqrt{3}$ and the proof is then follows.

Question 6: Can you decide which of the following two is greater?
$X= \sqrt{9}-\sqrt{8}+\sqrt{7}-\sqrt{6}+\sqrt{5}-\sqrt{4}+\sqrt{3}-\sqrt{2}$ or $Y=\sqrt{10}-\sqrt{9}+\sqrt{8}-\sqrt{7}+\sqrt{6}-\sqrt{5}+\sqrt{4}-\sqrt{3}$
A. $X\gt Y$.
B. $Y\gt X$.

From the facts that we've gathered from the above questions, we see that we can say the following inequalities hold:

$\sqrt{3}-\sqrt{2}\gt \sqrt{4}-\sqrt{3}$

$\sqrt{5}-\sqrt{4}\gt \sqrt{6}-\sqrt{5}$

$\sqrt{7}-\sqrt{6}\gt \sqrt{8}-\sqrt{7}$

$\sqrt{9}-\sqrt{8}\gt \sqrt{10}-\sqrt{9}$

Adding them up we get:

$\sqrt{9}-\sqrt{8}+\sqrt{7}-\sqrt{6}+\sqrt{5}-\sqrt{4}+\sqrt{3}-\sqrt{2}\gt \sqrt{10}-\sqrt{9}+\sqrt{8}-\sqrt{7}+\sqrt{6}-\sqrt{5}+\sqrt{4}-\sqrt{3}$

Therefore, we can conclude by now that $X\gt Y$ is true and hence A is the answer.