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Question of Lê Quốc Trần Anh
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Lê Quốc Trần Anh
Coodinator
13/06/2018 at 02:07
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Prove that with a,b,c > 0:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{a}\)
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