Cho x,y,z>0. x+y+z=1
Tìm Min P=\(\frac{xy}{x^4+y^4+xy}+\frac{yz}{y^4+z^4+yz}+\frac{xz}{x^4+z^4+xz}\)
Những câu hỏi liên quan
Cho x,y,z > 0 xy + yz +xz=1 tìm min của
\(\frac{1}{x^4-yz+2}\)+\(\frac{1}{y^4-xz+2}\)+ \(\frac{1}{z^4-xy+2}\)
Cho x, y,z >0. chứng minh:
\(\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}+\frac{\sqrt{xz}}{y+3\sqrt{yz}}\le\frac{3}{4}\)3/4
Đặt \(\left(x,y,z\right)\rightarrow\left(a,b,c\right)\) (chẳng có lý do j đâu mình gõ a,b,c quen hơn thôi)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(3P=\frac{3\sqrt{ab}}{c+3\sqrt{bc}}+\frac{3\sqrt{bc}}{a+3\sqrt{bc}}+\frac{3\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=3-\left(\frac{a}{a+3\sqrt{bc}}+\frac{b}{b+3\sqrt{ca}}+\frac{c}{c+3\sqrt{ab}}\right)\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a^2+b^2+c^2\right)+3\left(ab+bc+ca\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a^2+b^2+c^2\right)+\frac{\left(a+b+c\right)^2}{3}}\right]=3-\frac{9}{4}=\frac{3}{4}\)
Xảy ra khi \(a=b=c\)
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lý do đặt x,y,z= a,b,c
chỉ để copy nhanh hơn thôi :))
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Cho x,y,z thỏa mản :\(\frac{xy}{x+y}=\frac{12}{7},\frac{yz}{y+z}=-6,\frac{xz}{x+z}=-4\)-4 . Tìm x, y ,z
Cho x, y, z >0 thoả mãn x+y+z=1. Cmr: \(\frac{x}{x+yz}+\frac{y}{y+xz}+\frac{z}{z+xy}\le\frac{9}{4}\)
\(VT=\sum\frac{x}{x\left(x+y+z\right)+yz}=\sum\frac{x}{\left(x+y\right)\left(x+z\right)}=\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}\)
\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}-xyz}\)
\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+xyz-xyz}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
\(\frac{xy}{x^2+yz+xz}+\frac{yz}{y^2+xy+xz}+\frac{xz}{z^2+xy+yz}\le\frac{x^2+y^2+z^2}{xy+yz+xz}\)
cm biết x y z >0
#)Góp ý :
Mời bạn tham khảo :
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
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Tham khảo qua đây nè :
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%Ân-b%C3%ACnh-thu%E1%BA%ADn-2016-2017
tk cho mk nhé
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cho x,y,z>0 thỏa mãn x+y+z=3. Cmr:
\(\frac{2x^2+y^2+z^2}{4-yz}+\frac{2y^2+x^2+z^2}{4-xz}+\frac{2z^2+x^2+y^2}{4-xy}\ge4xyz\)
toán lớp mấy v
1hay 23456789
Cho x,y,z>0, x>y : xy+yz+xz+z2=1
Tìm Min:
\(P=\frac{1}{4\left(x-y\right)^2}+\frac{1}{\left(x+z\right)^2}+\frac{1}{\left(y+z\right)^2}\)
Cho các số dương x, y, z. CMR:
\(\frac{xy}{x^2+yz+xz}+\frac{yz}{y^2+xy+xz}+\frac{xz}{z^2+yz+xy}\le\frac{x^2+y^2+z^2}{xy+yz+xz}\)
Bunhiacopxki: \(\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)\ge\left(xy+yz+zx\right)^2\)
\(\Rightarrow\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Thiết lập tương tự và cộng lại:
\(\Rightarrow VT\le\frac{xy\left(y^2+yz+zx\right)+yz\left(z^2+xy+zx\right)+zx\left(x^2+yz+xy\right)}{\left(xy+yz+zx\right)^2}\)
\(VT\le\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\)
Ta chỉ cần chứng minh: \(\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\le\frac{x^2+y^2+z^2}{xy+yz+zx}\)
\(\Leftrightarrow xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz\le\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2yz+xy^2z+xyz^2\le x^3y+y^3z+z^3x\)
\(\Leftrightarrow\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\ge x+y+z\) (đúng theo Cauchy-Schwarz)
Dấu "=" xảy ra khi \(x=y=z\)
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Cho các số dương x,y,z . Chứng minh rằng:
\(\frac{xy}{x^2+yz+xz}+\frac{yz}{y^2+xy+xz}+\frac{xz}{z^2+yz+xy}\le\frac{x^2+y^2+z^2}{xy+yz+xz}\)
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
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