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Chứng minh a^2+b^2≥1/2

Cho a+b\(\ge\)1. CM:a2+b2\(\ge\dfrac{1}{2}\)

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Trả lời (1)

  • Áp dụng BĐT Cauchy-Schwarz ta có:

    \(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

    \(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

    \(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\ge1^2=1\)

    \(\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)

    Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)

      bởi Bee Bee HN 31/01/2019
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