Question

Cho a,b,c là các số dương thỏa mãn :a² +2b² < 3c².Chứng minh : \(\frac{1}{a}\)+\(\frac{2}{b}\)>\(\frac{3}{c}\)

Answer 1

ta có:\(\left(a+2b\right)^2=\left(1.a+\sqrt{2}.\sqrt{2}b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)\)( bđt bunhiacopxki)

\(\left(a+2b\right)^2\le3.3c^2=9c^2\)→\(a+2b\le3c\)

lại có:\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)

dấu = xảyra khi.... a+2b²=3c²(:v)