cho a,b,c>0; a+b+c<=1 tìm min p=a ²+b ²+c ²+1/a ²+1/b ²+1/c ²

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1. Đáp án: $ P\ge \dfrac{82}{3}$

   Giải thích các bước giải:

   Ta có:

   $P=a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$

   $\to P=a^2+b^2+c^2+\dfrac{1}{81}(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})+\dfrac{80}{81}(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})$

   $\to P\ge (a^2+b^2+c^2)+\dfrac{1}{81}\cdot \dfrac{9}{a^2+b^2+c^2}+\dfrac{80}{81}\cdot \dfrac13(\dfrac1a+\dfrac1b+\dfrac1c)^2$

   $\to P\ge (a^2+b^2+c^2)+ \dfrac{1}{9(a^2+b^2+c^2)}+\dfrac{80}{81}\cdot \dfrac13(\dfrac1a+\dfrac1b+\dfrac1c)^2$

   $\to P\ge 2\sqrt{(a^2+b^2+c^2)\cdot \dfrac{1}{9(a^2+b^2+c^2)}}+\dfrac{80}{81}\cdot \dfrac13(\dfrac{9}{a+b+c})^2$

   $\to P\ge \dfrac23+\dfrac{80}{81}\cdot \dfrac13(\dfrac{9}{1})^2$

   $\to P\ge \dfrac{82}{3}$

   Dấu = xảy ra khi $a=b=c=\dfrac13$

   Trả lời