(1) find all real values of x,y and z in
x^{2}-\left|x \right|=\left|yz \right|.......(1)
y^{2}-\left|y \right|=\left|zx \right|........(2)
z^{2}-\left|z \right|=\left|xy \right|........(3)
bhatt sir how can i solve these type of question.
341Let a=|x|, b=|y|, c=|z|
Then we have
\begin{matrix} a^2-a= bc\\ b^2-b = ac \\ c^2-c = ab \end{matrix}
Remember that a,b,c>0
Then we can write the inequalities
\begin{matrix} a^2\ge bc\\ b^2\ge ac \\ c^2 \ge ab \end{matrix}
WLOG a \ge b \ge c
If any of the inequalitiies here becomes a strict inequality, then c^2 \ge ab becomes absurd.
(or equivalently if either a2>bc or b2>ac or c2>ab then multiplying you get the absurdity (abc)^2 > abc
So we must have a =b=c =0
66Well there are more solutions: (±1, 0, 0), (0, ±1, 0), (0, 0, ±1)
341Continuing from Kaymant Sir's post:
That brings to the fore that some of the reasoning in #2 is wrong.
a=b=c = 0 is a trivial solution
As in #2, we have a^2 \ge bc; b^2 \ge ca ; c^2 \ge ab
If none of a,b,c are zero and one of the inequalities is a strict inequality, then on multiplication we get (abc)^2>a^2b^2c^2 which is absurd.
So we must have a^2=bc; b^2=ac;c^2=ab
which leads us to back a=b=c = 0
Now WLOG suppose a= 0. Then at least one of b or c is zero. So lets say b =0.
Then if c≠0, we must have c=1.
Hence (0,0,±1) and permutations are solutions
341Yeah, I checked up, the question is from that book, though he wouldnt have picked it up from there as solutions are given (it is pretty much on the same lines)
1708Thanks hsbhatt sir and kaymant sir. actually It is given in Mathematics Spectrum(Arihant Publication).
