Cho x,y,z là độ dài 3 cạnh của 1 tam giác. CMR:
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{3\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz}\ge9\)
cho x,y,z là số thực dương thỏa mãn xy+yz+xz=xyz
cmr \(\frac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\frac{yz}{x^3\left(1+y\right)\left(1+z\right)}+\frac{xz}{y^3\left(1+x\right)\left(1+z\right)}\ge\frac{1}{16}\)
Từ \(xy+yz+xz=xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a,b,c\right)\) thì có
\(\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(a+1\right)\left(c+1\right)}+\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{1}{16}\)\(\forall\hept{\begin{cases}a+b+c=1\\a,b,c>0\end{cases}}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{64}+\frac{c+1}{64}\ge\frac{3a}{16}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế
\(VT+\frac{2\left(a+b+c+3\right)}{64}\ge\frac{3\left(a+b+c\right)}{16}\Leftrightarrow VT\ge\frac{1}{16}\)
Khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=1\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
x;y;z>0. CMR: \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\ge2+\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
cho xyz=1.CMR
\(\frac{1}{x^3\left(y+z\right)}+\frac{1}{y^3\left(x+z\right)}+\frac{1}{z^3\left(y+z\right)}\ge\frac{3}{2}\)
đặt \(P=\frac{1}{x^3\left(y+z\right)}+\frac{1}{y^3\left(z+x\right)}+\frac{1}{z^3\left(x+y\right)}\)
\(=\frac{yz}{x^2\left(y+z\right)}+\frac{zx}{y^2\left(z+x\right)}+\frac{xy}{z^2\left(x+y\right)}\)
áp dụng bất đẳng thức cô si ta có:
\(\frac{yz}{x^2\left(y+z\right)}+\frac{y+z}{4yz}\ge\frac{1}{x};\frac{zx}{y^2\left(z+x\right)}+\frac{z+x}{4zx}\ge\frac{1}{y};\frac{xy}{z^2\left(x+y\right)}+\frac{x+y}{4xy}\ge\frac{1}{z}\)
\(\Rightarrow P+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Rightarrow P\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{2}.3\sqrt[3]{\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}=\frac{3}{2}\left(Q.E.D\right)\)
dấu bằng xảy ra khi x=y=z=1
cho x y z > 0 và xyz=1. Tìm Min của \(P=\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{y^3}{\left(1+y\right)\left(1+z\right)}+\frac{z^3}{\left(1+z\right)\left(1+x\right)}\)
dễ mà bạn :))) gáy tí , sai thì thôi
\(P=\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{y^3}{\left(1+y\right)\left(1+z\right)}+\frac{z^3}{\left(1+z\right)\left(1+x\right)}\)
\(=\frac{x^3\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}+\frac{y^3\left(1+x\right)}{\left(1+y\right)\left(1+x\right)\left(1+z\right)}+\frac{z^3\left(1+y\right)}{\left(1+x\right)\left(1+z\right)\left(1+y\right)}\)
\(=\frac{x^3\left(1+z\right)+y^3\left(1+x\right)+z^3\left(1+y\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\frac{3\sqrt[3]{x^3y^3z^3\left(1+x\right)\left(1+y\right)\left(1+z\right)}}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
đến đây áp dụng BĐT phụ ( 1+a ) ( 1+b ) ( 1+c ) >= 8abc
EZ :)))
nhưng làm thế thì ko bảo toàn đc dấu bất đẳng thức mà
TA LẦN LƯỢT ÁP DỤNG BĐT CAUCHY 3 SỐ VÀO TỪNG BDT SAU SẼ ĐƯỢC:
Có: \(\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge3\sqrt[3]{\frac{x^3\left(1+x\right)\left(1+y\right)}{64\left(1+x\right)\left(1+y\right)}}\)
=> \(\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge\frac{3x}{4}\)
CMTT TA CŨNG SẼ ĐƯỢC: \(\hept{\begin{cases}\frac{y^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge\frac{3z}{4}\end{cases}}\)
=> TA CỘNG TỪNG VẾ 3 BĐT ĐÓ LẠI SẼ ĐƯỢC:
\(\Rightarrow P+\frac{1+x}{4}+\frac{1+y}{4}+\frac{1+z}{4}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P+\frac{x+y+z+3}{4}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)-3}{4}\)
TA LẠI ÁP DỤNG BĐT CAUCHY 3 SỐ 1 LẦN NỮA SẼ ĐƯỢC:
\(\Rightarrow P\ge\frac{2.3\sqrt[3]{xyz}-3}{4}\)
\(\Rightarrow P\ge\frac{2.3-3}{4}=\frac{6-3}{4}=\frac{3}{4}\) (DO \(xyz=1\))
DẤU "=" XẢY RA <=> \(x=y=z\)
MÀ: \(xyz=1\Rightarrow x=y=z=1\)
VẬY P MIN \(=\frac{3}{4}\Leftrightarrow x=y=z=1\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
a , cho x,y,z >0 ; xyz =1
CMR: \(\frac{x^3}{\left(1+y\right).\left(1+z\right)}\)+\(\frac{y^3}{\left(1+z\right).\left(1+x\right)}\)+\(\frac{z^3}{\left(1+x\right).\left(1+y\right)}\ge\frac{3}{4}\)
cho 3 số dương x,y,z thỏa mãn : \(x+y+z=xyz\)
CMR : \(\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Khi đó \(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có: \(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra \(VT=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Đpcm
x,y,z>0,xyz=1
P=\(\frac{x^3}{\left(1+z\right)\left(1+y\right)}+\frac{y^3}{\left(1+x\right)\left(1+z\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\). tìm gtnn của P.
Cho \(x,y,z\) là các số thực dương. CMR: \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\ge2+\frac{2\left(x+y+z\right)}{3\sqrt{xyz}}\)
\(VT=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(=2+\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\)
Bài toán trở thành \(\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\ge\frac{x+y+z}{3\sqrt{xyz}}\)
Áp dụng bất đẳng thức AM-GM:
\(\frac{z}{x}+\frac{z}{y}+\frac{z}{z}\ge3\sqrt[3]{\frac{z^3}{xyz}}=\frac{3z}{\sqrt[3]{xyz}}\)
Tương tự:
\(\frac{y}{x}+\frac{y}{z}+\frac{y}{y}\ge\frac{3y}{\sqrt[3]{xyz}}\)
\(\frac{x}{z}+\frac{x}{y}+\frac{x}{x}\ge\frac{3x}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow VT+3\ge3+\frac{3}{\sqrt[3]{xyz}}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{3\left(x+y+z\right)}{\sqrt[3]{xyz}}\)\(\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Is it true?