cho x,y > 0 biết \(\dfrac{4}{x}\) + \(\dfrac{25}{y}\) = 1. Cmr: x + y ≥ 49
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DH
1
3 tháng 9 2022
\(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{1}{x+y}:\dfrac{1}{4}=\dfrac{4}{x+y}\)
\(\Rightarrow\left(x+y\right)^2>=4xy\)
\(\Leftrightarrow\left(x-y\right)^2>=0\)(luôn đúng)
AH
Akai Haruma
Giáo viên
14 tháng 10 2018
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+x+y+z)\geq (1+1+1+1)^2\)
\(\Rightarrow \frac{2}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{16}{2x+y+z}\)
Hoàn toàn tương tự:
\(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\geq \frac{16}{x+2y+z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\geq \frac{16}{x+y+2z}\)
Cộng theo vế các BĐT vừa thu được:
\(\Rightarrow 4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\Rightarrow 16\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\Rightarrow \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq 1\)
Ta có đpcm.
14 tháng 10 2018
Ta có :
\(\dfrac{1}{2x+y+z}=\dfrac{16}{16\left(x+x+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+2y+z}=\dfrac{16}{16\left(x+y+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}=\dfrac{16}{16\left(x+y+z+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
Cộng từng vế của BĐT ta được :
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Vậy BĐT đã được chứng minh !
TA
6 tháng 11 2018
\(\dfrac{1}{2x+y+z}=\dfrac{1}{x+y+x+z}\le\dfrac{1}{4}.\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\)
\(\le\dfrac{1}{4}.\dfrac{1}{4}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)=\dfrac{1}{16}.\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Tuong tu : \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}.\left(\dfrac{2}{y}+\dfrac{1}{z}+\dfrac{1}{x}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}.\left(\dfrac{2}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
=> \(VT\le\dfrac{1}{16}.\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{2}{y}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{2}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
= \(\dfrac{1}{16}.\left[4.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=1\left(dpcm\right)\)
6 tháng 11 2018
Áp dụng bđt Cauchy-Schwarz:
\(\dfrac{1}{2x+y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
Cộng theo vế suy ra đpcm. \("="\Leftrightarrow x=y=z=\dfrac{3}{4}\)
15 tháng 6 2018
Bài 1. Ta có : \(xy+\dfrac{1}{xy}=16xy-15xy+\dfrac{1}{xy}\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(x+y\) ≥ \(2\sqrt{xy}\)
⇔ \(\left(x+y\right)^2\) ≥ \(4xy\)
⇔ \(\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\) ≥ xy
⇔ - 15xy ≥ \(\dfrac{1}{4}.\left(-15\right)=\dfrac{-15}{4}\)
CMTT , \(16xy+\dfrac{1}{xy}\) ≥ \(2\sqrt{16xy.\dfrac{1}{xy}}=2.\sqrt{16}=8\)
⇒ \(16xy+\dfrac{1}{xy}\) - 15xy ≥ \(8-\dfrac{15}{4}=\dfrac{17}{4}\)
21 tháng 9 2017
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 \(a=\dfrac{1}{6};b=\dfrac{1}{3};c=\dfrac{1}{2}\)
UK
21 tháng 9 2017
2) \(\dfrac{1}{x}+\dfrac{25}{y}+\dfrac{64}{z}=\dfrac{4}{4x}+\dfrac{225}{9y}+\dfrac{1024}{16z}\ge\dfrac{\left(2+15+32\right)^2}{4x+9y+6z}=49\)
ND
1 tháng 4 2018
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\right)\ge6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\Leftrightarrow\dfrac{2x^2}{y^2}+\dfrac{2y^2}{x^2}+8\ge\dfrac{6x}{y}+\dfrac{6y}{x}\)
\(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}\right)-4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-4.\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)^2+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0^{\left(1\right)}\)
\(^{\left(1\right)}\)đúng \(\Rightarrowđpcm\)
1 tháng 4 2018
Áp dụng BĐT : x4 + y4 ≥ 2x2y2
=> \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) ≥ 2 ( x , y > 0 )
TT , \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2 ( x , y > 0 )
Ta có : \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) + 4 ≥ 6 ( 1 )
\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\) ≥ 6 ( 2 )
Từ ( 1 ; 2) => đpcm
NK
1
13 tháng 4 2017
\(BDT\Leftrightarrow\dfrac{x^4}{x^2y^2}+\dfrac{y^4}{x^2y^2}+\dfrac{4x^2y^2}{x^2y^2}\ge3\left(\dfrac{x^2}{xy}+\dfrac{y^2}{xy}\right)\)
\(\Leftrightarrow\dfrac{x^4+y^4-2x^2y^2+6x^2y^2}{x^2y^2}\ge\dfrac{3\left(x^2+y^2\right)}{xy}\)
\(\Leftrightarrow\dfrac{x^4+y^4-2x^2y^2}{x^2y^2}\ge\dfrac{3x^2+3y^2}{xy}-\dfrac{6xy}{xy}\)
\(\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2}{x^2y^2}\ge\dfrac{3\left(x^2-2xy+y^2\right)}{xy}=\dfrac{3\left(x-y\right)^2}{xy}\)
\(\Leftrightarrow\left(x-y\right)^2\left[\dfrac{\left(x+y\right)^2-3xy}{x^2y^2}\right]\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{x^2+y^2-xy}{x^2y^2}\right)\ge0\) (luôn đúng)
Vậy BĐT đã được chứng minh
Ta có:
\(1=\dfrac{2^2}{x}+\dfrac{5^2}{y}\ge\dfrac{\left(2+5\right)^2}{x+y}=\dfrac{49}{x+y}\)
\(\Rightarrow x+y\ge49\) (đpcm)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(14;35\right)\)