xét hiệu
\(\left(a^2+b^2\right)^2-ab\left(a+b\right)^2\ge0\)
=> \(a^4-a^3b+b^4-ab^3\ge0\)
=>\(\left(a^4-a^3b\right)-\left(ab^3-b^4\right)\ge0\)
=>\(a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
=>\(\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)luôn đúng
=> đpcm
CM CÁC BẤT ĐẲNG THỨC SAU
A) \(2\left(A^2+B^2\right)\ge\left(A+B\right)^2\ge2\left(AB+BA\right)\)
B) \(3\left(A^2+B^2+C^2\right)\ge\left(A+B+C\right)^2\ge3\left(AB+BC+CA\right)\)
Chứng minh:a)\(x^2+5x-3\ge\dfrac{-37}{4}\)
b)\(a^2+b^2+c^2\ge ab+bc+ac\)
c)\(8\left(x+\dfrac{1}{2}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
Cho a, b, c > 0 . CMR:
\(\frac{1}{a+b+c}\ge\frac{a^3}{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}+\frac{b^3}{\left(2b^2+c^2\right)\left(2b^2+a^2\right)}+\frac{c^3}{\left(2c^2+a^2\right)\left(2c^2+a^2\right)}\)
CM : \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\)
CMR: (\(\left(\dfrac{a+b}{2}+\dfrac{c+d}{2}\right)^2\ge\left(a+c\right)\left(b+d\right)\)
CMR: \(\left(a^2+b^2\right)\left(a^4+b^4\right)\) ≥ \(\left(a^3+b^3\right)^2\)
CMR: \(2\left(a^3+b^3\right)\) ≥ \(\left(a+b\right)\left(a^2+b^2\right)\) với a, b > 0
Chứng minh rằng:
a, \(a^2\)+\(b^2-2ab\ge0\)
b,\(\dfrac{a^2+b^2}{2}\ge ab\)
c,\(a\left(a+2\right)< \left(a+1\right)^2\)
d,\(m^2+n^2+2\ge2\left(m+n\right)\)
e, \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)(với a > 0, b > 0)
CMR: \(\left(\dfrac{a+b}{2}+\dfrac{c+d}{2}\right)\ge\left(a+c\right)\left(b+d\right)\)
