$A=\sum \frac{a}{\sqrt{a^2+8bc}}=\sum \frac{a^2}{a\sqrt{a^2+8bc}}$Áp dụng BĐT $Cauchy - schawrz$$A \geq \frac{(a+b+c)^2}{\sum a\sqrt{a^2+8bc}}$Ta có : đặt $P=a\sqrt{a^2+8bc}$$=> P^2 \leq (a+b+c)(a^3+b^3+c^3+24abc)$dễ dàng chứng minh được $a^3+b^3+c^3 +24abc \leq (a+b+c)^3$$=> P^2 \leq (a+b+c)^4$$=> P \leq (a+b+c)^2$$=> A \geq 1$
$A=\sum \frac{a}{\sqrt{a^2+8bc}}=\sum \frac{a^2}{a\sqrt{a^2+8bc}}$Áp dụng BĐT $Cauchy - schawrz$$A \geq \frac{(a+b+c)^2}{\sum a\sqrt{a^2+8bc}}$Ta có : đặt $P=a\sqrt{a^2+8bc}$$=> P^2 \leq (a+b+c)(a^3+b^3+c^3+24abc)$
$A=\sum \frac{a}{\sqrt{a^2+8bc}}=\sum \frac{a^2}{a\sqrt{a^2+8bc}}$Áp dụng BĐT $Cauchy - schawrz$$A \geq \frac{(a+b+c)^2}{\sum a\sqrt{a^2+8bc}}$Ta có : đặt $P=a\sqrt{a^2+8bc}$$=> P^2 \leq (a+b+c)(a^3+b^3+c^3+24abc)$
dễ dàng chứng minh được $a^3+b^3+c^3 +24abc \leq (a+b+c)^3$$=> P^2 \leq (a+b+c)^4$$=> P \leq (a+b+c)^2$$=> A \geq 1$