CAT 2024 Slot 1 · Quant · Question 8

The signs are easier to handle after a substitution: put u=x, v=-y and w=-z.

Then {x}^{2}+{y}^{2}+{z}^{2}={u}^{2}+{v}^{2}+{w}^{2} and x-y-z=u+v+w.

The two conditions become 4\left({u}^{2}+{v}^{2}+{w}^{2}\right)=a and 4\left( u+v+w \right)=3+a.

For any three reals, {\left( u+v+w \right)}^{2}\le 3\left({u}^{2}+{v}^{2}+{w}^{2}\right), by Cauchy-Schwarz.

Substituting, {\left( \frac{3+a}{4}\right)}^{2}\le 3\times \frac{a}{4}.

Multiplying by 16, {\left( 3+a \right)}^{2}\le 12a, that is 9+6a+{a}^{2}\le 12a.

Rearranging, {a}^{2}-6a+9\le 0, which is {\left( a-3 \right)}^{2}\le 0.

A square is never negative, so the only possibility is a=3.

This is attainable: a=3 forces equality in Cauchy-Schwarz, so u=v=w=\frac{1}{2}, that is x=\frac{1}{2} and y=z=-\frac{1}{2}.

Shortcut. Two equations in three unknowns normally leave a free parameter, so when a unique answer is demanded an inequality must be doing the work. Reaching for Cauchy-Schwaz and landing on a perfect square \le 0 is the standard finish.