2 replies to this topic

[tiendungthachthat]

Cho a;b;c>0 ; a+b+c=3. Tìm GTNN của 

M=$\frac{1}{a^{2}+1}+ \frac{1}{b^{2}+1} +\frac{1}{c^{2}+1}$

[tritanngo99]

Cho a;b;c>0 ; a+b+c=3. Tìm GTNN của 

M=$\frac{1}{a^{2}+1}+ \frac{1}{b^{2}+1} +\frac{1}{c^{2}+1}$

Ta có: $\frac{1}{a^2+1}\ge \frac{-1}{2}a+1\iff a(a-1)^2\ge 0(TRUE)\implies \sum \frac{1}{a^2+1}\ge \frac{-1}{2}\sum a+3=\frac{3}{2}.$

Edited by tritanngo99, 30-10-2016 - 10:45.

[Viet Hoang 99]

$\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge 1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}$

Tương tự

$M\ge 3-\dfrac{a+b+c}{2}=\dfrac{3}{2}$