Bài toán 2:Cho các số thực không âm$a,b,c$. Chứng minh rằng:
$a\sqrt{b^2+4c^2}+b\sqrt{4a^2+c^2}+c\sqrt{a^2+4b^2}\leq$$\frac{3}{4}(a+b+c)^2$
k mất tính tq gs $c=min\left \{ a,b,c \right \}$
$\left ( \sum _{cyc}a\sqrt{b^2+4c^2} \right )^2\leq \left ( \sum _{cyc}a(3a+b+5c) \right )\left ( \sum _{cyc}\frac{a(b^2+4c^2)}{3a+b+5c} \right )$
do đó ta chỉ cần c/m $\sum _{cyc}\frac{a(b^2+4c^2)}{3a+b+5c} \leq \frac{3}{16}\left ( \sum a\right )^2$
$\Leftrightarrow 45\sum a^5+165\sum _{cyc}a^4b+69\sum _{cyc}ab^4+536\sum a^3bc-306\sum _{cyc}a^3b^2-18\sum _{cyc}a^2b^3-410\sum a^2b^2c\geq 0$
$\Leftrightarrow 3(a+5b)(15a^2+10ab+3b^2)(a-b)^2+Ac\geq 0$
trong đó $A=69a^4+(536b-18c)a^3-(410b^2+410bc+306c^2)a^2+(536b^3-410b^2c+436bc^2+165c^3)a+165b^4-306b^3c-18b^2c^2+69bc^3+45c^4$
sd $c=min\left \{ a,b,c \right \}$ dễ dàng c/m đc $A\geq 0$