\(choA=\frac{457}{1}+\frac{456}{2}+\frac{455}{3}+...+\frac{1}{457}\) chứng minh A> 2016
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NK
1
6 tháng 3 2023
\(A=\left(\dfrac{456}{2}+1\right)+...+\left(\dfrac{2}{456}+1\right)+\left(\dfrac{1}{457}+1\right)+1\)
\(A=458+\dfrac{458}{2}+....+\dfrac{458}{456}+\dfrac{458}{457}-\dfrac{458}{458}\)
\(A=458\left(\dfrac{1}{2}+...+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\right)\)
Ta xét \(\dfrac{1}{2}+....+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\)có :
\(\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{8}>\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{1}{2}\)
\(\dfrac{1}{9}+\dfrac{1}{10}+....+\dfrac{1}{16}>\dfrac{1}{16}+....+\dfrac{1}{16}=\dfrac{1}{2}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+....+\dfrac{1}{32}>\dfrac{1}{32}+.....+\dfrac{1}{32}=\dfrac{1}{2}\)
\(\dfrac{1}{33}+\dfrac{1}{34}+....+\dfrac{1}{64}>\dfrac{1}{64}+....+\dfrac{1}{64}=\dfrac{1}{2}\)
\(\dfrac{1}{65}+\dfrac{1}{66}+.....+\dfrac{1}{128}>\dfrac{1}{128}+....+\dfrac{1}{128}=\dfrac{1}{2}\)
\(\dfrac{1}{129}+\dfrac{1}{130}+.....+\dfrac{1}{256}>\dfrac{1}{256}+....+\dfrac{1}{256}=\dfrac{1}{2}\)
\(\dfrac{1}{257}+\dfrac{1}{258}+....+\dfrac{1}{458}>\dfrac{1}{458}+...+\dfrac{1}{458}=\dfrac{1}{2}\)
Vậy ta thấy được rằng
\(\dfrac{1}{2}+...+\dfrac{1}{456}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{202}{458}\)
\(=4+\dfrac{202}{458}=\dfrac{2034}{458}\)
Vậy \(A>458.\dfrac{2034}{458}=2034\)
Hay tức là A > 2016 ( đpcm )
MP
23 tháng 3 2022
Ta có:
A = (457/1 + 1) + (456/2 + 1) + ... + (2/456 + 1) + (1/457 + 1) - 457
A = 458 + 458/2 + ... + 458/456 + 458/457 - 457
A = 458 (1 + 1/2 + ...+ 1/456 + 1/457) - 457
Xét 1 + 1/2 + ... + 1/456 + 1/457, ta có
1 = 1
1/2 = 1/2
1/3 + 1/4 > 1/4 + 1/4 = 1/2
1/5 + 1/6 + ... + 1/8 > 1/8 + 1/8 + ... + 1/8 = 1/2
1/9 + 1/10 +...+ 1/16 > 1/16 + 1/16 +...+ 1/16 = 1/2
1/17 + 1/18 + ... + 1/32 > 1/32 + ... + 1/32 = 1/2
1/33+ 1/34 + ... + 1/64 > 1/64 + ...+ 1/64 = 1/2
1/65 + 1/66 + ...+ 1/128 > 1/128 + ... + 1/128 = 1/2
1/129 + 1/130 + ... + 1/256 > 1/256 + ...+ 1/256 = 1/2
1/257 + 1/258 + ... + 1/457 > 1/457 + ... + 1/457 = 201/457 > 0,4
Vậy 1 + 1/2 + ... + 1/456 + 1/457 > 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 0,4 = 5,4
Vậy A > 458*5,4 - 457 = 2016,2
Vậy A > 2016.
MP
23 tháng 3 2022
Ta có:
A = (456/2 + 1) + ... + (2/456 + 1) + (1/457 + 1) + 1
A = 458 + 458/2 + ... + 458/456 + 458/457 - 458/458
A = 458 (1/2 + ...+ 1/456 + 1/457 + 1/458)
Xét 1/2 + ... + 1/456 + 1/458, ta có
1/2 = 1/2
1/3 + 1/4 > 1/4 + 1/4 = 1/2
1/5 + 1/6 + ... + 1/8 > 1/8 + 1/8 + ... + 1/8 = 1/2
1/9 + 1/10 +...+ 1/16 > 1/16 + 1/16 +...+ 1/16 = 1/2
1/17 + 1/18 + ... + 1/32 > 1/32 + ... + 1/32 = 1/2
1/33+ 1/34 + ... + 1/64 > 1/64 + ...+ 1/64 = 1/2
1/65 + 1/66 + ...+ 1/128 > 1/128 + ... + 1/128 = 1/2
1/129 + 1/130 + ... + 1/256 > 1/256 + ...+ 1/256 = 1/2
1/257 + 1/258 + ... + 1/458 > 1/458 + ... + 1/458 = 202/458
Vậy 1/2 + ... + 1/456 + 1/457 > 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 202/458 = 4 + 202/458 = 2034/458
Vậy A > 458*2034/458 = 2034
Vậy A > 2016.
29 tháng 11 2016
\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}=\frac{a^4}{ab+ac}+\frac{b^4}{ba+bc}+\frac{c^4}{ca+cb}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\)
24 tháng 9 2018
\(a^2+b^2+c^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(=\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2+\left(a+\frac{1}{a}\right)+\left(b+\frac{1}{b}\right)+\left(c+\frac{1}{c}\right)+\left(a+b+c\right)-3\)
\(\ge2+2+2+3-3=6\)
5 tháng 2 2018
Bài này dễ,ông không chịu làm thì có ^_^:
Ta có:\(B=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+....+\left(\frac{1}{2^{2014}+1}+....+\frac{1}{2^{2015}}\right)+\frac{1}{2^{2015}+1}+...+\frac{1}{2^{2016}-1}\)
\(>1+\frac{1}{2}+2.\frac{1}{2^2}+2^2.\frac{1}{2^3}+........+2^{2014}.\frac{1}{2^{2015}}\)
\(=1+\frac{1}{2}+\frac{1}{2}+.........+\frac{1}{2}\) (có 2015 phân số \(\frac{1}{2}\))
\(=1+2014.\frac{1}{2}+\frac{1}{2}=1008+\frac{1}{2}>1008\)
17 tháng 8 2017
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)-9abc\ge0\)
\(\Leftrightarrow a^2b+a^2c+abc+abc+ab^2+b^2c+abc+ac^2+bc^2-9abc\ge0\)
\(\Leftrightarrow a^2b+a^2c+ab^2+b^2c+ac^2+bc^2-6abc\ge0\)
\(\Leftrightarrow\left(a^2b-2abc+bc^2\right)+\left(a^2c-2abc+b^2c\right)+\left(ab^2-2abc+ac^2\right)\ge0\)
\(\Leftrightarrow b\left(a-b\right)^2+c\left(a-c\right)^2+a\left(b-c\right)^2\ge0\)(luôn đúng \(\forall a;b;c>0\))
Vật bđt đã đc chứng minh
17 tháng 8 2017
Cho a,b,c>0 thì dễ thôi :v
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}\)
Khi a=b=c
T
28 tháng 7 2019
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\) (do a,b,c >0)
Ta có đpcm
ta có: A=457/1+456/2+455/3+...+1/457
suy ra: A= 457/1+1+456/2+1+...+1/457+1-467
suy ra: A=458/1+458/2+458/3+...+458/457 -457
suy ra A=458(1+1/2+1/3+...+1/457)-457
suy ra :A=458 * 6,702992749 -457
suy ra: A>2291>2016
suy ra: A>2016
vậy A>2016