Year: 2008
Paper: 1
Question Number: 3
Course: LFM Pure
Section: Proof
There were significantly more candidates attempting this paper this year (an increase of nearly 25%), but many found it to be very difficult and only achieved low scores. The mean score was significantly lower than last year, although a similar number of candidates achieved very high marks. This may be, in part, due to the phenomenon of students in the Lower Sixth (Year 12) being entered for the examination before attempting papers II and III in the Upper Sixth. This is a questionable practice, as while students have enough technical knowledge to answer the STEP I questions at this stage, they often still lack the mathematical maturity to be able to apply their knowledge to these challenging problems. Again, a key difficulty experienced by most candidates was a lack of the algebraic skill required by the questions. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many students were simply unable to progress on some questions because they did not know how to handle the algebra. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was also pleasing that one of the applied questions, question 13, attracted a very large number of attempts. However, the examiners were again left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers and other available resources to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Prove that, if $c\ge a$ and $d\ge b$, then
\[
ab+cd\ge bc+ad\,.
\tag{$*$}
\]
\begin{questionparts}
\item If $x\ge y$, use $(*)$ to show that $x^2+y^2\ge 2xy\,$.
If, further, $x\ge z$ and $y\ge z$, use $(*)$ to show that $z^2+xy\ge xz+yz$ and deduce that $x^2+y^2+z^2\ge xy+yz+zx\,$.
Prove that the inequality $x^2+y^2+z^2\ge xy+yz+zx\,$ holds for all $x$, $y$ and $z$.
\item Show similarly that the inequality
\[\frac st +\frac tr +\frac rs
+\frac ts +\frac rt +\frac sr
\ge 6\]
holds for all positive $r$, $s$ and $t$.
\end{questionparts}\begin{align*}
&& \underbrace{(c-a)}_{\geq 0}\underbrace{(d-b)}_{\geq 0} & \geq 0 \\
\Leftrightarrow && cd -bc -ad + ab &\geq 0 \\
\Leftrightarrow && ab +cd &\geq bc+ad \\
\end{align*}
\begin{questionparts}
\item Applying $(*)$ with $c=d=x$ and $a=b=y$ we obtain: $x^2 + y^2 \geq xy + xy = 2xy$
Similarly, applying $(*)$ with $c=x, d=y, a=b=z$ we obtain: $z^2 + xy \geq zx+zy$ so
$x^2+y^2+z^2 \geq 2xy + z^2 \geq xy + zx+zy$
There was nothing special about our choice of ordering $x,y,z$ so it is true for all $x,y,z$
\item \begin{align*}
\frac st +\frac tr +\frac rs
+\frac ts +\frac rt +\frac sr &=\left ( \frac st+\frac ts \right)+\left ( \frac tr +\frac rt \right)+\left ( \frac rs
+\frac sr \right) \\
& \geq 2 \sqrt{\frac st \frac ts} + 2 \sqrt{\frac tr \frac rt} + 2 \sqrt{\frac rs \frac sr} \\
& = 2 + 2 + 2 \\
&= 6
\end{align*}
\end{questionparts}
This was another popular question, although the scores were again fairly poor. The proof of (*) was often done quite well. The main difficulties here arose because of a lack of clarity in the logic; it is important to make clear where the starting point is and what steps are being taken to move forward from there. A significant number of candidates attempted to work backwards, and then divided by d − b or the like without realising that this might be zero. Also, inequalities were multiplied without any regard to the sign of the numbers under consideration; for example, while 0 > −1 and 1 > −2, it is not true that 0 × 1 > (−1) × (−2). The beginning of part (i) was completed correctly by a majority of candidates. It is important to stress again that if a question specifies "use (*)", then this must be done to gain any credit; no marks were given for the numerous answers which began with "as (x − y)² ≥ 0 for all x and y, we have x² − 2xy + y² ≥ 0" or similar. The last part of (i) was often poorly tackled. It was sometimes interpreted to mean "when x < 0" or other spurious cases, without understanding that the inequality had so far only been shown in the case x ≥ y ≥ z. (Indeed, the intermediate result z² + xy ≥ xz + yz does not hold in the case x > z > y.) Many other candidates ignored their preceding work and went on to prove the result from scratch using the inequality (x − y)² + (y − z)² + (z − x)² ≥ 0. Very few candidates explained the symmetry of the situation. Part (ii) was problematic because of the wording of the question. It turned out that there is a very straightforward way to answer this part by making use of the results proved in part (i). While this was not what was actually asked ("Show similarly . . . "), it was felt unfair to penalise candidates too harshly for taking this route. Thus they were awarded partial credit and all such candidates were referred to the Chief Examiner for individual consideration. Nonetheless, the attempts at this part, by whichever method, were generally either close to perfect or non-starters.