Ta có: $b + c = (b + c).(a + b + c)^2$ (vì $a + b + c = 1$) Ta có $[ a + (b + c) ]^2 \geq 4(b+c)a$ (vì $(x + y)^2 \geq 4xy )$ $\Leftrightarrow (b + c).(a + b + c)^2\geq 4(b+c)^2.a$ lại có $(b+c)^2\geq 4bc \Rightarrow 4(b+c)^2.a \geq 16abc$ (đpcm)b+c=(b+c).(a+b+c)2
Ta có: $b + c = (b + c).(a + b + c)^2$ (vì $a + b + c = 1$) Ta có $[ (a + b) + c ]^2 \geq 4(a + b)c$ (vì $(x + y)^2 \geq 4xy )$ $\Leftrightarrow (b + c).(a + b + c)^2\geq 4(a + b)^2.c$ lại có $(a + b)^2\geq 4ab \Rightarrow 4(a + b)^2.c \geq 16abc$ (đpcm)b+c=(b+c).(a+b+c)2
Ta có: $b + c = (b + c).(a + b + c)^2$ (vì $a + b + c = 1$) Ta có $[ a +
(b + c
) ]^2 \geq 4(b
+c
)a$ (vì $(x + y)^2 \geq 4xy )$ $\Leftrightarrow (b + c).(a + b + c)^2\geq 4(
b+
c)^2.
a$
lại có $(
b+
c)^2\geq 4b
c \Rightarrow 4(b
+c)^2.
a \geq 16abc$ (đpcm)b+c=(b+c).(a+b+c)2