show that a^2+d^2=b^2+c^2
Proof:
by Cauchy-Schawartz in equality we have
(a^2+b^2+c^2+d^2)(1^2 + \sqrt2^2+ \sqrt3^2+2^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2
Or 10(a^2 + b^2 + c^2+ d^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2
Or \sqrt{10(a^2 + b^2+ c^2+d^2)}>=(a+b\sqrt{2}+c\sqrt{3}+2d)
from given condition and above we have
\sqrt{10(a^2 + b^2+c^2+d^2)}=(a+b\sqrt{2}+c\sqrt{3}+2d)^2
when
\frac{a}{1}= \frac{b}{\sqrt2}=\frac{c}{\sqrt3}= \frac{d}{2}= k (say)
So a = k, b^2= 2k^2,c^2=3k^2,d=2k and hence a^2+d^2=b^2+c^2
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