T
trungphongfc


Cho a,b,c \geq $\frac{-1}{4}$ Chứng minh rằng
$\sqrt[2]{4a+1}$ + $\sqrt[2]{4b+1}$ +$\sqrt[2]{4c+1}$ <= $\sqrt[2]{21}$
$\sqrt[2]{4a+1}$ + $\sqrt[2]{4b+1}$ +$\sqrt[2]{4c+1}$ <= $\sqrt[2]{21}$
Last edited by a moderator: