Question

1)a,b,c >0 ; a+b+c=1. CMR:

\(\dfrac{a}{1+a}\) + \(\dfrac{2b}{2+b}\) +\(\dfrac{3c}{c+3}\) \(\le\) \(\dfrac{6}{7}\)

2) x,y,z >0; 4x+9y+16z=49

CMR: \(\dfrac{1}{x}\) + \(\dfrac{25}{y}\) + \(\dfrac{64}{z}\) \(\ge\) 49

Answer 1

Từ \(\dfrac{a}{1+a}+\dfrac{2b}{2+b}+\dfrac{3c}{3+c}\le\dfrac{6}{7}\)

\(\Leftrightarrow1-\dfrac{a}{1+a}+2-\dfrac{2b}{2+b}+3-\dfrac{3c}{3+c}\ge6-\dfrac{6}{7}\)

\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\ge\dfrac{36}{7}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\)

\(\ge\dfrac{\left(1+2+3\right)^2}{a+b+c+6}=\dfrac{36}{7}=VP\)

Xảy ra khi \(a=\dfrac{1}{6};b=\dfrac{1}{3};c=\dfrac{1}{2}\)

Answer 2

2) \(\dfrac{1}{x}+\dfrac{25}{y}+\dfrac{64}{z}=\dfrac{4}{4x}+\dfrac{225}{9y}+\dfrac{1024}{16z}\ge\dfrac{\left(2+15+32\right)^2}{4x+9y+6z}=49\)