Cho a+b+c=3 .chứng minh $\frac{a^{2}}{a+b^{2}}$ +$\frac{b^{2}}{b+c^{2}}$ +$\frac{c^{2}}{c+a^{2}}$ $\geq$ $\frac{3}{2}$ gấp vô cùngggg

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1. Giải thích các bước giải:

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

   Ta có:

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

   $\to\dfrac{a^2}{a+b^2}\ge \dfrac{2a^2}{a^2+2b^2+1}$

   Tương tự ta có:

   $\dfrac{b^2}{b+c^2}\ge \dfrac{2b^2}{b^2+2c^2+1}$

   $\dfrac{c^2}{c+a^2}\ge \dfrac{2c^2}{c^2+2a^2+1}$

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

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

   $\to P\ge 2(\dfrac{a^4}{a^4+2a^2b^2+a^2}+\dfrac{b^4}{b^4+2b^2c^2+b^2}+\dfrac{c^4}{c^4+2c^2a^2+c^2})$

   $\to P\ge 2\cdot \dfrac{(a^2+b^2+c^2)^2}{a^4+2a^2b^2+a^2+b^4+2b^2c^2+b^2+c^4+2c^2a^2+c^2}$

   $\to P\ge 2\cdot \dfrac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2+(a^2+b^2+c^2)}$

   $\to P\ge 2\cdot \dfrac{9(a^2+b^2+c^2)^2}{9(a^2+b^2+c^2)^2+9(a^2+b^2+c^2)}$

   $\to P\ge 2\cdot \dfrac{9(a^2+b^2+c^2)^2}{9(a^2+b^2+c^2)^2+(a+b+c)^2(a^2+b^2+c^2)}$

   $\to P\ge 2\cdot \dfrac{9(a^2+b^2+c^2)^2}{9(a^2+b^2+c^2)^2+3(a^2+b^2+c^2)(a^2+b^2+c^2)}$

   $\to P\ge 2\cdot \dfrac{9(a^2+b^2+c^2)^2}{12(a^2+b^2+c^2)^2}$

   $\to P\ge \dfrac32$

   Dấu = xảy ra khi $a=b=c=1$

   Trả lời