Dimostrare $(a^2+b^2)(p^2+q^2)\ge (ap+bq)^2$

Text

Prove that, for every choice of real numbers $a,b,p,q$, $(a^2+b^2)(p^2+q^2)\ge (ap+bq)^2.$

Solution

Expand the difference between the left and right sides: $(a^2+b^2)(p^2+q^2)-(ap+bq)^2=a^2p^2+a^2q^2+b^2p^2+b^2q^2-\big(a^2p^2+2apbq+b^2q^2\big).$ After simplification the terms $a^2q^2-2apbq+b^2p^2=(aq-bp)^2\ge 0$ remain. Hence the left side is $\ge$ the right side. Equality holds if and only if $aq=bp$. $\boxed{(a^2+b^2)(p^2+q^2)\ge (ap+bq)^2,\qquad =\ \text{iff } aq=bp.}$