Cho cac so duong x y z t co tong bang 1
CMR \(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}\ge64\)
Những câu hỏi liên quan
cho x,y,z la cac so thuc thoa x+y+z=0, x+1>0, y+1>0, z+1>0. tim GTLN cua P=\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}\)
cho x,y,z,t la cac so duong. tim GTNN cua A=\(\frac{x-t}{t+y}+\frac{t-y}{y+z}+\frac{y-z}{z+x}+\frac{z-x}{x+t}\)
cho x,y,z la cac so duong thoa man \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)
CMR:\(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\le1\)
Áp dụng AM-GM ta có \(\frac{1^2}{x}+\frac{1^2}{x}+\frac{1^2}{y}+\frac{1^2}{z}\ge\frac{\left(1+1+1+1\right)^2}{2x+y+z}\)
hay \(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\)
Tương tự : \(\frac{2}{y}+\frac{1}{x}+\frac{1}{z}\ge\frac{16}{2y+x+z}\) ; \(\frac{2}{z}+\frac{1}{x}+\frac{1}{y}\ge\frac{16}{2z+x+y}\)
Cộng theo vế : \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge16\left(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\right)\)
\(\Leftrightarrow\)\(16\left(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\right)\le16\)
\(\Leftrightarrow\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\le1\)
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Cho các số dương x, y, z, t có tổng bằng 1. Chứng minh rằng:
\(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}\)\(\ge64\)
Áp dụng bđt Cauchy schwarz:
=> 1/x+1/y+4/z+16/t >= [(1+1+2+4)^2] / x+y+z+t=8^2/(x+y+z+t)=64/1=64
=> đpcm.
Áp dụng BĐT Svac - xơ:
\(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}\ge\frac{\left(1+1+2+4\right)^2}{x+y+z+t}=\frac{64}{1}=64\)
(Dấu "="\(\Leftrightarrow x=y=\frac{1}{22};z=\frac{2}{11};t=\frac{8}{11}\))
Sửa)):
(Dấu "="\(\Leftrightarrow x=y=\frac{1}{16};z=\frac{1}{4};t=1\))
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Voi cac so duong, CMR:
\(\frac{1}{x+3y}+\frac{1}{y+3z}+\frac{1}{z+3x}\ge\frac{1}{x+2y+z}+\frac{1}{y+2z+x}+\frac{1}{z+2x+y}.;\)
1.Cho x,y,z,t >0
CMR : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\le\frac{16}{x+y+z+t}\)
cau 1: Cho A frac{100^{2014}+2}{3}-frac{100^{2015}+17}{9}.tong cac cua so cua B-9ACau 2: So sanh Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+........+frac{1}{100.101}voi 1 ta doc A....1Cau 3 : Cho bon so a,b,c,d sao cho a+b+c+d khac 0 . Biet frac{b+c+d}{a}frac{c+d+a}{b}frac{d+a+b}{c}frac{a+b+c}{d}k. Vay kCau 4 : so cac so nguyen am x thoa man x^{2015}left(-2right)^{2014}Cau 5; tim x,y,z biet frac{x}{y}frac{10}{9},frac{y}{z}frac{3}{4}va x-y+z78Cau 6: tap hop cac so co ba chu so chia het cho 18 va to...
Đọc tiếp
cau 1: Cho A= \(\frac{100^{2014}+2}{3}-\frac{100^{2015}+17}{9}.\)tong cac cua so cua B=-9A
Cau 2: So sanh A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{100.101}\)voi 1 ta doc A....1
Cau 3 : Cho bon so a,b,c,d sao cho a+b+c+d khac 0 . Biet \(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}=k\). Vay k=
Cau 4 : so cac so nguyen am x thoa man \(x^{2015}=\left(-2\right)^{2014}\)
Cau 5; tim x,y,z biet \(\frac{x}{y}=\frac{10}{9},\frac{y}{z}=\frac{3}{4}\)va x-y+z=78
Cau 6: tap hop cac so co ba chu so chia het cho 18 va tong cac chu so ti le voi 1;2;3 la
Cau 7: gia tri cua tong S=1.2+2.3+.....+49.50 la S=
Cho x,y,z,t >0
C/m : \(\left(x+y+z+t\right)\left(\frac{1}{x+y+z}+\frac{1}{y+z+t}+\frac{1}{z+t+x}+\frac{1}{t+x+y}\right)\ge\frac{16}{3}\)
bạn dùng BĐT Cauchuy-Swartch cho cs Bt thứ 2 là ra nhé
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Cho hai so duong x,y co tong bang 1
Tim GTNN cua P=\(\left(1-\frac{1}{x^2}\right)\left(1-\frac{1}{y^2}\right)\)
Cho x,y,z la cac so thuc duong thoa man xyz=2
Chung minh rang:\(\frac{x}{2x^2+y^2+5}+\frac{2y}{6y^2+z^2+6}+\frac{4z}{3z^2+4x^2+16}\le\frac{1}{2}\)