Gerard Sayson TMUA Mock Paper 1 · Question 18

A student is to prove the following theorem:

For any positive real numbers $a,b,c\in\mathbb R$ , if $a^2+b^2+c^2=1$ then $a+b+c\le\sqrt3$ .

The student uses some of the statements below to construct a direct deductive proof of the above theorem. They are not necessarily arranged in the correct order.

I. Substitute $a^2+b^2+c^2=1$ into the inequality to get $(a+b+c)^2\le3$ .
II. Assume $a+b+c\le\sqrt3$ . Since $a,b,c>0$ , squaring both sides gives $(a+b+c)^2\le3$ .
III. We have $ab+bc+ca\le a^2+b^2+c^2$ .
IV. We have $(a+b+c)^2\le3(a^2+b^2+c^2)$ .
V. Since $a,b,c>0$ , have $a+b+c>0$ . Take the positive square root to get $a+b+c\le\sqrt3$ .
VI. For all real numbers $a,b,c$ , have $(a-b)^2+(b-c)^2+(c-a)^2\ge0$ .
VII. Therefore, $3(a^2+b^2+c^2)\ge(a+b+c)^2$ , proving the theorem.

What is the correct flow of statements that the student needs?

\mathrm{VI}\to\mathrm{III}\to\mathrm{IV}\to\mathrm{I}\to\mathrm{V}

\mathrm{VI}\to\mathrm{IV}\to\mathrm{III}\to\mathrm{I}\to\mathrm{V}

\mathrm{II}\to\mathrm{I}\to\mathrm{IV}\to\mathrm{III}\to\mathrm{VI}

\mathrm{II}\to\mathrm{VII}\to\mathrm{VI}\to\mathrm{III}\to\mathrm{IV}

\mathrm{I}\to\mathrm{V}\to\mathrm{VI}\to\mathrm{III}\to\mathrm{IV}

examday

Gerard Sayson TMUA Mock Paper 1 · Q18