Ta có: $a^2+2b^2+3=a^2+b^2+b^2+1+2$
Áp dụng bất đẳng thức Cosi ta được:
$a^2+b^2≥2ab;b^2+1≥2b$
$⇒a^2+b^2+b^2+1≥2ab+2b$
$⇒a^2+b^2+b^2+1+2≥2ab+2b+2=2(ab+b+1)$
Hay $a^2+2b^2+3≥2(ab+b+1)$
$⇒\dfrac{1}{a^2+2b^2+3}≤\dfrac{1}{2(ab+b+1)}$
Chứng minh tương tự ta có:
$\dfrac{1}{b^2+2c^2+3}≤\dfrac{1}{2(bc+c+1)}$
$\dfrac{1}{c^2+2a^2+3}≤\dfrac{1}{2(ca+a+1)}$
$⇒\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}≤\dfrac{1}{2(ab+b+1)}+\dfrac{1}{2(bc+c+1)}+\dfrac{1}{2(ca+a+1)}=\dfrac{1}{2}.(\dfrac{1}{(ab+b+1)}+\dfrac{1}{(bc+c+1)}+\dfrac{1}{(ca+a+1)})$
Ta có: $\dfrac{1}{(ab+b+1)}+\dfrac{1}{(bc+c+1)}+\dfrac{1}{(ca+a+1)}$
$=\dfrac{abc}{b(a+1+ac)}+\dfrac{abc}{c(b+1+ab)}+\dfrac{1}{ca+a+1}$
$=\dfrac{ac}{ac+a+1}+\dfrac{ab}{b(a+ac+1)}+\dfrac{1}{ac+a+1}$
$=\dfrac{ac+a+1}{ac+a+1}=1$
Nên $\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}≤\dfrac{1}{2}.(\dfrac{1}{(ab+b+1)}+\dfrac{1}{(bc+c+1)}+\dfrac{1}{(ca+a+1)})=\dfrac{1}{2}$
