Cho \(x;y;z>0\)
\(CMR:\) \(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\le\frac{3}{4}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng bất đẳng thức bu nhi a ta có
\(\left(a+2b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)=3.\left(a^2+2b^2\right)\le3.3c^2=9c^2\)
=> \(a+2b\le3c\)
Mà \(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)
=> \(\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\left(ĐPCM\right)\)
Ta có:
\(2M=\frac{2ab}{a+b+2}=\frac{\left(a+b\right)^2-\left(a^2+b^2\right)}{a+b+2}\)
\(=\frac{\left(a+b\right)^2-4}{a+b+2}=a+b-2\le\sqrt{2\left(a^2+b^2\right)}-2\)
\(=2\sqrt{2}-2\)
\(\Rightarrow M\le\sqrt{2}-1\)
Ta có :
\(2M=\frac{2ab}{a+b+2}\)
\(=\frac{\left(a+b\right)^2-\left(a^2+b^2\right)}{a+b+2}\)
\(=\frac{\left(a+b\right)^2-4}{a+b+2}\)
\(\Leftrightarrow a+b-2\le\sqrt{2\left(a^2+b^2\right)}-2\)
\(=2\sqrt{2}-2\)
\(\Leftrightarrow M\le\sqrt{2}-1\)
a) \(\left(x+3\right)\left(x-1\right)-2\left(x+3\right)^2+\left(x-4\right)\left(x+4\right)\)
\(=x^2-x+3x-3-2\left(x^2+6x+9\right)+x^2-16\)
\(=2x^2+2x-19-2x^2-12x-18\)
\(=-10x-37\)
b) \(\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{\left(5^2-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{5^{32}-1}{24}\)
a) (x+3)(x-1)-2(x+302)+(x-4)(x+4)=x2+2x-3-2x-1800+x2-16=2x2-1819
b)...=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)/(5^2-1)=(5^4-1)(5^4+1)(5^8+1)(5^16+1)/(5^2-1)
=(5^8-1)(5^8+1)(5^16+1)/(5^2-1)=(5^16-1)(5^16+1)/(5^2-1)=(5^32-1)/(5^2-1)
????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????how to?
Có : (a-b)^2>=0
<=> a^2+b^2-2ab >=0
<=>a^2+b^2 >= 2ab
<=>a^2+b^2+2ab >= 4ab
<=> (a+b)^2 >= 4ab
Với a,b >0 thì chia cả 2 vế cho (a+b).ab thì :
a+b/ab >= 4/a+b
<=>4/a+b <= 1/a+1/b
<=> 1/a+b <= 1/4.(1/a+1/b) ( với mọi a,b > 0 )
Áp dụng bđt trên cho x;y;z > 0 thì : x/2x+y+z = x. 1/(x+y)+(z+x) <= x/4 .( 1/x+y+1/x+z) = x/4.(x+y) + x/4.(x+z)
Tương tự : y/x+2y+z <= y/4.(y+x) + y/4.(y+z)
z/x+y+2z <= z/4.(z+x) + z/4.(z+y)
=> VT <= [ x/4.(x+y) + y/4.(y+x) ] + [ y/4.(y+z) + z/4.(z+y) ] + [ z/4.(z+x) + x/4.(x+z) ] = 1/4 + 1/4 + 1/4 = 3/4
=> ĐPCM
Dấu "=" xảy ra <=> x=y=z > 0
k mk nha
áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) với mọi a,b >0
Thì \(\frac{x}{x+y}+\frac{x}{x+z}\ge\frac{4x}{2x+y+z}\)
Tương tự thì đpcm
Cách này nhanh này thành đơ