Cho `x,y,x >0` thoả mãn `sum_{cyc} (x+y)^2 <= 4` `CMR: sum_{cyc} {x^2+1}/{(y+z)^2}>=3`

1 bình luận về “Cho `x,y,x >0` thoả mãn `sum_{cyc} (x+y)^2 <= 4` `CMR: sum_{cyc} {x^2+1}/{(y+z)^2}>=3`”

1. Ta có (x^2+1)/(y+z)^2+(y^2+1)/(x+z)^2+(z^2+1)/(y+x)^2

       =(x^2)/(y+z)^2+1/(y+z)^2+(y^2)/(x+z)^2+1/(x+z)^2+(z^2)/(y+x)^2+1/(y+x)^2

   Áp dụng bđt phụ 3(a^2+b^2+c^2)ge(a+b+c)^2 ta có

   3(sum_(cyc)(x^2)/(y+z)^2)ge(sum_(cyc)x/(y+z))^2

   ⇔3(sum_(cyc)(x^2)/(y+z)^2)ge(3/2)^2

   ⇔sum_(cyc)(x^2)/(y+z)^2ge3/4

   Áp dụng bđt Cauchy-schwarz

   sum_(cyc)1/(y+z)^2ge9/(sum_(cyc)(x+y)^2)ge9/4

   ⇒sum_(cyc)(x^2+1)/(y+z)^2ge9/4+3/4=3(đpcm)

   Dấu = xảy ra khi x=y=z=1/sqrt{3}

    

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