Giải thích các bước giải:
Ta có:
$\dfrac{a+1}{b^2+1}=\dfrac{(a+1)(b^2+1)-(a+1)b^2}{b^2+1}=(a+1)-\dfrac{(a+1)b^2}{b^2+1}$
$\to \dfrac{a+1}{b^2+1}=(a+1)-\dfrac{(a+1)b^2}{\dfrac{b^2}{4}+\dfrac{b^2}{4}+\dfrac{b^2}{4}+\dfrac{b^2}{4}+1}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac{(a+1)b^2}{5\sqrt[5]{\dfrac{b^2}{4}\cdot\dfrac{b^2}{4}\cdot\dfrac{b^2}{4}\cdot\dfrac{b^2}{4}\cdot1}}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac{(a+1)b^2}{5\sqrt[5]{\dfrac{b^8}{4^4}}}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac{(a+1)b^2}{5\sqrt[5]{\dfrac{b^{10}\cdot 4}{b^2\cdot 4^5}}}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac{(a+1)b^2}{5\cdot \dfrac{b^2}{4}\cdot \sqrt[5]{\dfrac{4}{b^2}}}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac45\cdot \dfrac{a+1}{\sqrt[5]{\dfrac{4}{b^2}}}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac45\cdot (a+1)\cdot\sqrt[5]{\dfrac{b^2}{4}}$
Mà $\dfrac{b}{2}+\dfrac{b}{2}+1+1+1\ge 5\sqrt[5]{\dfrac{b}{2}\cdot \dfrac{b}{2}\cdot 1\cdot 1\cdot 1}$
$\to b+3\ge 5\sqrt[5]{\dfrac{b^2}{4}}$
$\to \dfrac15(b+3)\ge \sqrt[5]{\dfrac{b^2}{4}}$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac45\cdot (a+1)\cdot\dfrac15(b+3)$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac4{25}\cdot (a+1)\cdot(b+3)$
$\to \dfrac{a+1}{b^2+1}\ge (a+1)-\dfrac4{25}\cdot (ab+3a+b+3)$
Tương tự chứng minh được:
$\dfrac{b+1}{c^2+1}\ge (b+1)-\dfrac4{25}\cdot (bc+3b+c+3)$
$\dfrac{c+1}{a^2+1}\ge (c+1)-\dfrac4{25}\cdot (ca+3c+a+3)$
Cộng vế với vế ba bất đẳng thức trên ta được:
$VT\ge (a+b+c+3)-\dfrac4{25}(ab+bc+ca+4(a+b+c)+9)$
$\to VT\ge (a+b+c+3)-\dfrac4{25}(\dfrac13(a+b+c)^2+4(a+b+c)+9)$
$\to VT\ge (6+3)-\dfrac4{25}(\dfrac13\cdot 6^2+4\cdot 6+9)$
$\to VT\ge \dfrac95$
Dấu = xảy ra khi $a=b=c=2$
$\to đpcm$
