So I have the following problem, which I'm having trouble solving:
Let a1 , a2 , ... , an be real numbers. Let b1 , b2 , ... , bn be positive real numbers. Prove
a21b1+a22b2+⋅⋅⋅+a2nbn≥(a1+a2+⋅⋅⋅+an)2b1+b2+⋅⋅⋅+bn
I was thinking that I somehow could use the Cauchy–Schwarz inequality, but with no success.
Any help would be very appreciated
Answer
You can simply use the cauchy-scwartz on the sets {a1√b1,a2√b2,…,an√bn} and {√b1,√b2,…,√bn}.
What you will get, is (a21b1+a22b2+⋅⋅⋅+a2nbn)(b1+b2+⋯+bn)≥(a1+a2+⋯+an)2.
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