\(\frac{4}{x^2+7}=\frac{4}{x^2+1+y^2+1+z^2+1+x^2+1}\le\frac{4}{4x+2y+2z}=\frac{2}{2x+y+z}\)
đến đây tự làm nha
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cho \(x^2+y^2+z^2=3\)
C/m \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{4}{x^2+7}+\frac{4}{y^2+7}+\frac{4}{z^2+7}\)
cho x;y;z là các số thực dương thỏa mãn x+y+z=1.CMR:
\(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{zx}{z^2+x^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{15}{4}\)
a)Cho các số x,y,z ge1.CMR: frac{1}{1+x}+frac{1}{1+y}+frac{1}{1+z}gefrac{3}{1+sqrt[3]{xyz}}.b) Cho x,y,z ge0 và xle1;yle1;zle1chứng minh:frac{1}{1+x^2}+frac{1}{1+y^2}+frac{1}{1+z^2}lefrac{3}{1+xyz}c)Cho a + bge2.CMR: a^3+b^3le a^4+b^4d)Cho a2+b2gefrac{1}{4}.CMR:a^4+b^4gefrac{1}{32}
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a)Cho các số x,y,z \(\ge\)1.CMR: \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{3}{1+\sqrt[3]{xyz}}\).
b) Cho x,y,z \(\ge\)0 và x\(\le1;y\le1;z\le1\)chứng minh:
\(\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}\le\frac{3}{1+xyz}\)
c)Cho a + b\(\ge\)2.CMR: \(a^3+b^3\le a^4+b^4\)
d)Cho a2+b2\(\ge\frac{1}{4}.CMR:a^4+b^4\ge\frac{1}{32}\)
CMR:
\(\frac{x^4}{y^2\left(x+z\right)}+\frac{y^4}{z^2\left(x+y\right)}+\frac{z^4}{x^2\left(y+z\right)}\ge\frac{x+y+z}{2}\)
cho x , y , z > 0 thỏa mãn xy + yz + zx = 3xyz
CMR: \(A=\frac{x^3}{z+x^2}+\frac{y^3}{x+y^2}+\frac{z^3}{y+z^2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
CHO a,b,c0 thỏa mãn: a^2b^2+b^2c^2+c^2a^2ge a^2+b^2+c^2CMR: frac{a^2b^2}{c^3left(a^2+b^2right)}+frac{b^2c^2}{a^3left(b^2+c^2right)}+frac{c^2a^2}{b^3left(a^2+c^2right)}gefrac{sqrt{3}}{2}ĐẶT Afrac{a^2b^2}{c^3left(a^2+b^2right)}+frac{b^2c^2}{a^3left(b^2+c^2right)}+frac{c^2a^2}{b^3left(c^2+a^2right)}ĐẶT:frac{1}{a}x,frac{1}{y}b,frac{1}{z}cRightarrow x^2+y^2+z^2ge1Rightarrow Afrac{x^3}{y^2+z^2}+frac{y^3}{z^2+x^2}+frac{z^3}{z^2+y^2}TA CÓ:xleft(y^2+z^2right)frac{1}{sqrt{2}}sqrt{2x^2left(y^2+z^2right)lef...
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CHO a,b,c>0 thỏa mãn: \(a^2b^2+b^2c^2+c^2a^2\ge a^2+b^2+c^2\)
CMR: \(\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(a^2+c^2\right)}\ge\frac{\sqrt{3}}{2}\)
ĐẶT \(A=\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(c^2+a^2\right)}\)
ĐẶT:\(\frac{1}{a}=x,\frac{1}{y}=b,\frac{1}{z}=c\)
\(\Rightarrow x^2+y^2+z^2\ge1\)
\(\Rightarrow A=\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{z^2+y^2}\)
TA CÓ:
\(x\left(y^2+z^2\right)=\frac{1}{\sqrt{2}}\sqrt{2x^2\left(y^2+z^2\right)\left(y^2+z^2\right)}\le\frac{1}{\sqrt{2}}\sqrt{\frac{\left(2x^2+2y^2+2z^2\right)^3}{27}}=\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)TƯƠNG TỰ:
\(y\left(x^2+z^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2},z\left(x^2+y^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)LẠI CÓ:
\(A=\frac{x^3}{y^2+z^2}+\frac{y^3}{x^2+z^2}+\frac{z^3}{x^2+y^2}=\frac{x^4}{x\left(y^2+z^2\right)}+\frac{y^4}{y\left(x^2+z^2\right)}+\frac{z^4}{z\left(x^2+y^2\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)}\ge\frac{1}{3.\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}}
\)\(\ge\frac{\sqrt{3}}{2}\sqrt{x^2+y^2+z^2}\ge\frac{\sqrt{3}}{2}\)
DẤU BẰNG XẢY RA\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\Rightarrow DPCM\)
cho các số thực dương x;y;z thỏa mãn \(\frac{1}{x^2+2}+\frac{1}{y^2+2}+\frac{1}{z^2+2}=\frac{1}{3}.\)CMR:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{3}{\sqrt{7}}\)
Cho các số thực dương a,b,c thỏa mãn x+y+z=1 .Chứng minh
\(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{zx}{z^2+x^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{15}{4}\)
cho x+y+z4 cmr frac{1}{xy}+frac{1}{yz}ge1BLTA CẦN CM frac{1}{x}left(frac{1}{y}+frac{1}{z}right)ge1Leftrightarrowfrac{1}{y}+frac{1}{z}ge xmà x4-left(y+zright)Rightarrowfrac{1}{y}+frac{1}{z}ge4-left(y+zright)Leftrightarrowfrac{1}{y}-2+y+frac{1}{z}-2+zge0Leftrightarrowleft(frac{1}{sqrt{y}}-sqrt{y}right)^2+left(frac{1}{sqrt{z}}-sqrt{z}right)^2ge0(luôn đúng)
Đọc tiếp
cho x+y+z=4
cmr \(\frac{1}{xy}+\frac{1}{yz}\ge1\)
BL
TA CẦN CM \(\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge1\Leftrightarrow\frac{1}{y}+\frac{1}{z}\ge x\)
mà x=\(4-\left(y+z\right)\)
\(\Rightarrow\frac{1}{y}+\frac{1}{z}\ge4-\left(y+z\right)\Leftrightarrow\frac{1}{y}-2+y+\frac{1}{z}-2+z\ge0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{y}}-\sqrt{y}\right)^2+\left(\frac{1}{\sqrt{z}}-\sqrt{z}\right)^2\ge0\)(luôn đúng)