giups minh voi cac ban nha. Kho qua hu hu: Cho a>0, b>0 va a+b=1 chung minh rang \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2>=12,5\)
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Cho a,b,c,d>0 va abcd=1 Chung minh
\(\frac{1}{\left(1+a\right)^2}\)+\(\frac{1}{\left(1+b\right)^2}\)+\(\frac{1}{\left(1+c\right)^2}\)+\(\frac{1}{\left(1+d\right)^2}\)>1
GIUP MK VOI CAC BAN OI
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cho a khac 0 b khac 0 va a+b=1 chung minh rang \(\frac{b}{a^3-1}-\frac{a}{b^3-1}=\frac{2\left(a-b\right)}{a^2b^2+3}\)
Cho các số thực dương a,b,c,d. Chung minh rang \(\frac{b}{\left(a+\sqrt{b}\right)^2}+\frac{a}{\left(b+\sqrt{a}\right)^2}\ge\frac{\sqrt{bd}}{ac+\sqrt{bd}}\)
Giup mk voi cac ban
cho a, b, c la cac so thuc duong thoa man a + b + c =abc chung minh rang :
\(\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ac\right)}+\frac{1}{c^2\left(1+ab\right)}\le\frac{1}{4}\)
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
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1.tìm các nghiem nguyen cua phuong trinh: 54x^3+1=y^3
2.cho x+y=1 và xy khac 0.chung mih \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
3.cho a,b,c la cac so thuc duong.chung minh :\(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)^2+\frac{14abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
Câu 2 thế y = 1 - x rồi quy đồng như bình thường là ra bn nhé
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Cho a,b,c > 0
Chung minh rang : \(\frac{\left(2b+3c\right)^2}{a}+\frac{\left(2c+3a\right)^2}{b}+\frac{\left(2a+3b\right)^2}{c}\ge25\left(a+b+c\right)\)
Áp dụng bất đẳng thức Cauchy–Schwarz dạng Engel ta có :
\(VT\ge\frac{\left(2b+3c+2c+3a+2a+3b\right)^2}{a+b+c}\)
\(=\frac{\left(5a+5b+5c\right)^2}{a+b+c}=\frac{\left[5\left(a+b+c\right)\right]^2}{a+b+c}\)
\(=\frac{25\left(a+b+c\right)^2}{a+b+c}=25\left(a+b+c\right)=VP\)
=> đpcm
Đẳng thức xảy ra <=> a = b = c
CMR \(\left(\frac{a+b+c}{3}\right)^3>=abc\)voi a,b,c>0
giup minh di minh chuan bi di hoc roi. nhanh len giup minh di HU HU.
1.xho x+y=1 và xy khác 0.chung minh \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
2.cho a,b,c là các số thực dương.chứng minh \(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2+\frac{14abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
bai 1 cho a,b,c>0
CMR: \(\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}>=1\)
cac ban oi giup minh. minh dang can gap lam. . lam on giup minh di. hu hu