Cho a,b,c >0 t/m: a^2+2b^2 =< 3c^2
chứng minh: 1/2 +2/b >= 3/c
[tex]a^2 +2b^2 \leq 3c^2 [/tex] chứng minh [tex] \frac{1}{2}+ \frac{2}{b}\geq \frac{3}{c}[/tex]
[TEX] (a+2b)^2=(1.a+\sqrt{2}.\sqrt{2b})^2 \leq (1+2)(a^2+2b^2)\leq 3.3c^2=9c^2[/TEX]
---> [TEX]a+2b \leq3c[/TEX]
[TEX]\frac{1}{a}+\frac{2}{b} =\frac{1}{a}+\frac{1}{b}+\frac{1}{b} \geq \frac{(1+1+1)^2}{a+b+b}= \frac{9}{a+2b} \geq \frac{9}{3c}=\frac{3}{c}[/TEX]