Đặt \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\) ta cần chứng minh
\(x^4+2y^4+3z^4\ge6\left(\dfrac{x+2y+3z}{6}\right)^4\)
Ta có:
\(x^4+2y^4+3z^4\ge\dfrac{\left(x^2+2y^2+3z^2\right)^2}{6}\ge\dfrac{1}{6}.\dfrac{\left(x+2y+3z\right)^4}{6^2}=6\left(\dfrac{x+2y+3z}{6}\right)^4\)
cho a,b,c>0, CMR:
\(\left(a+b+\dfrac{1}{4}\right)^2+\left(b+c+\dfrac{1}{4}\right)^2+\left(c+a+\dfrac{1}{4}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
Cho a,b,c>0.Chứng minh:
\(a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{b}+\dfrac{1}{a}\right)\ge6\)
Cho a,b,c CMR
\(a,a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)
\(b,a^4+b^4\ge\dfrac{\left(a+b\right)^4}{8}\)
\(c,a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\)
\(d,a^4+b^4+c^4\ge\dfrac{\left(a+b+c\right)^4}{27}\)
MÌNH CẦN GẤP GIÚP MÌNH NHA
Cho \(\dfrac{\sin^4\alpha}{a}+\dfrac{\cos^4\alpha}{b}=\dfrac{1}{a+b}\).CM:\(\dfrac{\sin^8\alpha}{a^3}+\dfrac{cos^8\beta}{b^3}=\dfrac{1}{\left(a+b\right)^3}\)
a,Rút gọn :
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)...\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\left(6^4+\dfrac{1}{4}\right)...\left(52^4+\dfrac{1}{4}\right)}\)
b, Tìm nghiệm nguyên: \(4x^2-8y^3+2z^2+4x-4=0\)
Cho a,b,c >0 và abc=1. Tìm min:
\(P=\dfrac{a^4+b^4}{\left(a^2+b^2\right)\left(a+b\right)}+\dfrac{b^4+c^4}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{a^4+c^4}{\left(a^2+c^2\right)\left(a+c\right)}\)
Cho a,b,c>0 thỏa mãn a+b+c=3 CMR:
\(\dfrac{a^4}{\left(a+2\right)\left(b+2\right)}+\dfrac{b^4}{\left(b+2\right)\left(c+2\right)}+\dfrac{c^4}{\left(c+2\right)\left(a+2\right)}\ge\dfrac{1}{3}\)
Tính các tích sau:
P\(_1\) =\(\left(1+\dfrac{2}{4}\right)\left(1+\dfrac{2}{10}\right)\left(1+\dfrac{2}{18}\right)....\left(1+\dfrac{2}{n^2+3n}\right)\)
P\(_2\) =\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{2}{n^2+2n}\right)\)
P\(_3\) = \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right).....\left(1-\dfrac{1}{1+2+3+4+...+n}\right)\)
P\(_4\) = \(\dfrac{2^4+4}{4^4+4}.\dfrac{6^4+4}{8^4+4}.\dfrac{8^4+4}{10^4+4}....\dfrac{18^4+4}{20^4+4}\)
Rút gọn các biểu thức
a, \(A=\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\dfrac{2+\sqrt{8}}{1+\sqrt{2}}\)
b, \(B=\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\) (với x >0, x ≠ 4)
