CMR : \(\left|x-1\right|+\left|x-5\right|\ge4\)
CMR :
a, \(\left|x-1\right|+\left|x-3\right|+\left|x-5\right|+\left|x-7\right|\ge8\)
b, \(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|\ge4\)
c, \(\left|x-1\right|+2\left|x-3\right|+\left|x-5\right|\ge4\)
Cần gấp lời giải đầy đủ dễ hiểu và đáp án nhé
\(\left|x-1\right|+\left|x-3\right|+\left|x-5\right|+\left|x-7\right|=\left(\left|x-1\right|+\left|x-7\right|\right)+\left(\left|x-3\right|+\left|x-5\right|\right)\\ \)
\(=\left(\left|x-1\right|+\left|7-x\right|\right)+\left(\left|x-3\right|+\left|5-x\right|\right)\)
\(\ge\left|x-1+7-x\right|+\left|x-3+5-x\right|=\left|6\right|+\left|2\right|=8\)
\(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|=\left(\left|x+1\right|+\left|x+3\right|\right)+\left|x+5\right|=\left(\left|x+1\right|+\left|3-x\right|\right)+\left|x+5\right|\)
\(\ge\left|x+1+3-x\right|+\left|x+5\right|=\left|4\right|+\left|x+5\right|=4+\left|x+5\right|\ge4\)
\(\left|x-1\right|+2\left|x-3\right|+\left|x-5\right|=\left(\left|x-1\right|+\left|x-5\right|\right)+2\left|x-3\right|=\left(\left|x-1\right|+\left|5-x\right|\right)+2\left|x-3\right|\)
\(\ge\left|x-1+5-x\right|+2\left|x-3\right|=\left|4\right|+2\left|x-3\right|=4+2\left|x-3\right|\ge4\)
Cho \(\left(x+1\right)\left(y+1\right)\)và \(xy=1\)
CMR: \(\left(x+1\right)\left(y+1\right)\ge4\)
Đề bài sai .
Nếu x = -1
y = -1
Cho \(x,y,z\ge0\) thỏa mãn \(x+y+z=1\) . CMR \(x+2y+z\ge4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)
Ta có: \(x+y+z=1\) nên:
\(\Rightarrow y+z=1-x\)
Thay \(y+z=1-x\) và áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) ta được:
\(4\left(1-x\right)\left(1-y\right)\left(1-z\right)=4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left[\left(y+z\right)+\left(1-z\right)\right]^2\left(1-y\right)\)
\(\Rightarrow4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left(1+y\right)^2\left(1-y\right)=\left(1+y\right)\left(1-y^2\right)\le1+y\)
\(\Rightarrow4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le1+y=x+2y+z\left(đpcm\right)\)
Cho x,y,z > 0 CMR
\(\frac{\left(y+z\right)^2}{x}+\frac{\left(x+z\right)^2}{y}+\frac{\left(x+y\right)^2}{z}\ge4\left(x+y+z\right)\)
Áp dụng BĐT cauchy schawrz dạng engel ta có:
\(\frac{\left(y+z\right)^2}{x}+\frac{\left(x+z\right)^2}{y}+\frac{\left(x+y\right)^2}{z}\ge\frac{\left(y+z+x+z+x+y\right)^2}{x+y+z}=\frac{4\left(x+y+z\right)^2}{x+y+z}=4\left(x+y+z\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Áp dụng BĐT cauchy schawrz dạng engel, ta có:
\(\frac{\left(y+z\right)^2}{x}+\frac{\left(x+z\right)^2}{y}+\frac{\left(x+y\right)^2}{z}\ge\frac{\left(y+z+x+z+x+y\right)^2}{x+y+z}=\frac{4\left(x+y+z\right)^2}{x+y+z}=4\left(x+y+z\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Áp dụng bất đẳng thức Svacxo ta có :
\(\frac{\left(y+z\right)^2}{x}+\frac{\left(x+z\right)^2}{y}+\frac{\left(x+y\right)^2}{z}\ge\frac{\left(y+z+x+z+x+y\right)^2}{x+y+z}\)
\(=\frac{\left(2x+2y+2z\right)^2}{x+y+z}=\frac{\left[2\left(x+y+z\right)\right]^2}{x+y+z}=\frac{4\left(x+y+z\right)^2}{x+y+z}=4\left(x+y+z\right)\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z\)
Vậy ta có điều phải chứng minh
Cho x; y; z > 0 sao cho (z+x)(z+y) = 1 CMR : \(\frac{1}{\left(x-y\right)^2}+\frac{1}{\left(z+x\right)^2}+\frac{1}{\left(z+y\right)^2}\ge4.\)
\(\hept{\begin{cases}x+z=a\\y+z=b\end{cases}}\); \(x-y=\left(x+z\right)-\left(y+z\right)=a-b\)
\(ab=1\Rightarrow b=\frac{1}{a}\)
\(A=VT=\frac{1}{\left(a-b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{\left(a-\frac{1}{a}\right)^2}+\frac{1}{a^2}+a^2\)
\(=\frac{a^2}{\left(a^2-1\right)^2}+a^2+\frac{1}{a^2}\)
\(t=a^2>0\)
\(A=\frac{t}{\left(t-1\right)^2}+t+\frac{1}{t}\)
\(A-4=\frac{\left(t^2-3t+1\right)^2}{t\left(t-1\right)^2}\ge0\)
\(\Rightarrow A\ge4\)
Dấu bằng xảy ra khi \(t=a^2=\frac{3\pm\sqrt{5}}{2}\)\(\Leftrightarrow a=\sqrt{\frac{3\pm\sqrt{5}}{2}}\)
\(\Leftrightarrow\hept{\begin{cases}a=x+z=\sqrt{\frac{3+\sqrt{5}}{2}}\\b=y+z=\sqrt{\frac{3-\sqrt{5}}{2}}\end{cases}}\) và hoán vị còn lại
Hệ trên có vô số nghiệm, chẳng hạn
\(\hept{\begin{cases}z=\frac{1}{10}\\x=\sqrt{\frac{3+\sqrt{5}}{2}}-\frac{1}{10}\\y=\sqrt{\frac{3-\sqrt{5}}{2}}-\frac{1}{10}\end{cases}}\)
Chào anh! Em mới học lớp 7 nên không biết làm. Nếu là toán lớp 9 thì anh nên đăng ký tài khoản ở h, sẽ có câu trả lời nhanh hơn đấy. Chúc anh học tốt!
Cho \(x,y,z>0\). CMR \(\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
\(VT=\dfrac{a+b}{2\sqrt[3]{abc}}+\dfrac{b+c}{2\sqrt[3]{abc}}+\dfrac{c+a}{2\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\) (AM-GM 4 số hạng)
Cho \(x,y,z\ge0\)T/M : \(x+y+z=1\)
CMR : \(x+2y+z\ge4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)
Áp dụng BĐT AM - GM, ta có:
\(4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)
\(=4\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
\(\le\frac{\left(x+y+y+z\right)^2}{4}\times4\left(x+z\right)\)
\(=\left(x+2y+z\right)^2\left(x+z\right)\)
\(\le\left(x+2y+z\right)\times\frac{\left(x+2y+z+x+z\right)^2}{4}\)
\(=\left(x+2y+z\right)\times\frac{4\left(x+y+z\right)^2}{4}\)
\(=x+2y+z\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c = \(\frac{1}{3}\)
Dấu = xảy ra:\(\hept{\begin{cases}x=z=\frac{1}{2}\\y=0\end{cases}}\)
cho các số thực không âm đôi một khác nhau thỏa mãn \(\left(x+z\right)\left(z+y\right)=1\)
Cmr: \(\frac{1}{\left(x-y\right)^2}+\frac{1}{\left(x+z\right)^2}+\frac{1}{\left(z+y\right)^2}\ge4\)
Cho x, y, z là các số thực dương. CMR:
\(\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)\ge4\left(xy+yz+zx\right)\)
Áp dụng BĐT Bunhiacopxki :
\(\left(x+y\right)\left(x+z\right)\ge\left(\sqrt{x}\sqrt{x}+\sqrt{y}\sqrt{z}\right)^2=\left(x+\sqrt{yz}\right)^2\)
\(\Rightarrow\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\)
Tương tự ta CM được:
\(\sqrt{\left(y+z\right)\left(y+x\right)}\ge y+\sqrt{xz}\) ; \(\sqrt{\left(x+z\right)\left(y+z\right)}\ge z+\sqrt{yx}\)
đặt vế trái của BĐT cần CM là A
\(\Rightarrow A=\left(x+y\right)\sqrt{\left(z+x\right)\left(z+y\right)}+\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}+\left(z+x\right)\sqrt{\left(y+z\right)\left(y+x\right)}\)
\(\ge\left(x+y\right)\left(z+\sqrt{xy}\right)+\left(y+z\right)\left(x+\sqrt{yz}\right)+\left(z+x\right)\left(y+\sqrt{zx}\right)\)
\(=\sqrt{xy}\left(x+y\right)+\sqrt{yz}\left(y+z\right)+\sqrt{zx}\left(z+x\right)+2\left(xy+yz+zx\right)\)
\(\ge2xy+2yz+2zx+2\left(xy+yz+zx\right)=4\left(xy+yz+zx\right)\)
Dấu ''='' xảy ra \(\Leftrightarrow x=y=z\)