Cho $a,b,c$ dương thoả $abc=1$. Tìm $Min$ của $A=\frac{a^{2}}{1+b}+\frac{b^{2}}{1+c}+\frac{c^{2}}{1+a}.$
Tìm Min: $A=\frac{a^{2}}{1+b}+\frac{b^{2}}{1+c}+\frac{c^{2}}{1+a}.$
Started By kunsomeone, 02-03-2016 - 19:57
#1
Posted 02-03-2016 - 19:57
#2
Posted 02-03-2016 - 20:10
Cho $a,b,c$ dương thoả $abc=1$. Tìm $Min$ của $A=\frac{a^{2}}{1+b}+\frac{b^{2}}{1+c}+\frac{c^{2}}{1+a}.$
$A\geq \frac{(a+b+c)^{2}}{3+a+b+c}$
Đặt a + b+c = t, $t\geq 3$
$\frac{t^{2}}{3+t}\geq \frac{3t-3}{4}\Leftrightarrow (t-3)^{2}\geq 0 \Rightarrow A\geq \frac{3.3-3}{4}=\frac{3}{2}$
Đẳng thức xảy ra <=> a=b=c=1
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