bai thi hoc ki kho day

T

tell_me_goobye

cho cac so thuc thoa man :
a + b + c = -2 va a^2 + b^2 + c^2 = 2
cmr: a, b, c [tex]\varepsilon[/tex] [-4/3;0]
minh can bai lam chi tiet
[TEX]\left{a+b+c=-2\\{a^2+b^2+c^2=2}\Leftrightarrow \left{b+c=-2-a \\{b^2+c^2=2-a^2[/TEX]
we have [TEX](b+c )^2+(b-c)^2 \ge\ (b+c)^2\Rightarrow 2(b^2+c^2)\ge\ (b+c)^2[/TEX]
\Rightarrow[TEX]2(2-a^2)\ge\ (-2-a )^2\Leftrightarrow 4-2a^2 \ge\ 4+4a+a^2[/TEX]
[TEX]\Leftrightarrow 3a^2+4a \le\ 0[/TEX]
\Leftrightarrow[TEX] \frac{-4}{3} \le\ a \le\ 0[/TEX]
simalar to what we have to prove
 
Top Bottom