Cho 3 số thực dương a,b,c thỏa mãn: [tex]a^2+b^2+c^2[/tex] =3.
Chứng minh:
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \frac{4}{a^2+7}+\frac{4}{b^2+7}+\frac{4}{c^2+7}$
[tex]\frac{1}{a+b}+\frac{1}{b+c}\geq \frac{4}{a+2b+c}(Cauchy-Schwarz)\\ \Rightarrow 2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq \frac{4}{a+2b+c}+\frac{4}{b+2c+a}+\frac{4}{c+2a+b}\\ \Rightarrow \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \frac{2}{a+2b+c}+\frac{2}{b+2c+a}+\frac{2}{c+2a+b}=\frac{4}{2a+4b+2c}+\frac{4}{2b+4c+2a}+\frac{4}{2c+4a+2b}[/tex]
BĐT cần chứng minh [tex]\Leftrightarrow \frac{4}{2a+4b+2c}+\frac{4}{2b+4c+2a}+\frac{4}{2c+4a+2b}\geq \frac{4}{a^2+7}+\frac{4}{b^2+7}+\frac{4}{c^2+7}\\[/tex]
Mà: [tex]2a+4b+2c\leq b^2+7<=>2a+4b+2c\leq b^2+a^2+b^2+c^2+4=a^2+2b^2+c^2+4\\ <=>(a-1)^2+2(b-1)^2+(c-1)^2\geq 0(luôn đúng)\\ \Rightarrow \frac{4}{2a+4b+2c}\geq \frac{4}{b^2+7}[/tex]
CMTT với 2 cái còn lại => đpcm
Dấu "=" xảy ra <=> a = b = c = 1