MHB Inequality involving a, b, c and d

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Given the real numbers $a,\,b,\,c$ and $d$, prove that

$(1+ab)^2+(1+cd)^2+a^2c^2+b^2d^2\ge 1$
 
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Expanding the LHS of the inequality we get $1+2ab+a^2b^2+1+2cd+c^2d^2+a^2c^2+b^2d^2=1+(1+ab+cd)^2+(ac-bd)^2\ge 1$
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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