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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: `a^2/(a^2+2abc)`+`b^2/(b^2+2abc`+`c^2/(c^2+2abc`$\geq$ `(a+b+c)^2/(a^2+b^2+c^2+6abc)`

Toán Lớp 8: a^2/(a^2+2abc)+b^2/(b^2+2abc+c^2/(c^2+2abc$\geq$ (a+b+c)^2/(a^2+b^2+c^2+6abc)

Comments ( 1 )

  1. $P=\frac{a^2}{a^2+2abc}+$ $\frac{b^2}{b^2+2abc}+$ $\frac{c^2}{c^2+2bc}$
    Theo bất đẳng thức Cauchy – Schwars (cộng mẫu)
    -> P ≥ $\frac{(a+b+c)^2}{a^2+b^2+c^2+2abc+2abc+2abc}$ 
    -> P ≥ $\frac{(a+b+c)^2}{a^2+b^2+c^2+6abc}$ (đpcm)
    Chúc bạn học tốt !!!

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222-9+11+12:2*14+14 = ? ( )