\(a^2+b^2-a-b-ab\ge-1\)
\(\Leftrightarrow2a^2+2b^2-2a-2b-2ab+2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(a-1\right)^2\ge0\)
\(\Rightarrow P\ge0\)
BĐT đúng
Given a set three integers greater than 1.
Let A be the number that's 1 less than the product of three given integers.
Let B be the product of numbers that're 1 less than three given integers.
Known that A is a multiple of B.
How many sets can you find.
A. 1
B. 2
C. 3
D. 4
Let # be defined by a#b=ab+a2+b2
Let @ be defined by a@b=\(\dfrac{a}{3}\)-b
Calculate (a#b)@a for a=15,b=6
Bài 1: Cho a,b,c∈ R. Chứng minh các bất đẳng thức sau:
a) \(ab\le\left(\frac{a+b}{2}\right)^2\le\frac{a^2+b^2}{2}\)
b) \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\) ; với a,b ≥ 0
c) a4+b4 ≥ a3b + ab3
d) a4+3 ≥ 4a
e) a3+b3+c3 ≥ 3abc ; với a,b,c > 0
f) \(a^4+b^4\le\frac{a^6}{b^2}+\frac{b^6}{a^2}\) ; với a,b ≠ 0
g) \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) ; với ab ≥ 1
h) (a5+b5)(a+b) ≥ (a4+b4)(a2+b2) ; với ab > 0
a) Cho a,b,c >0
Chứng minh: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
b) Cho a,b \(\ge\)1 , chứng minh:
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
Toán tiếng anh: Let a, b, c be the possible integer such that ab+bc=518 and ab-ac=360. Find the largest possible value of the product abc
Áp dụng bất đẳng thức cosi chứng minh
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với a,b \(\ge\)0
\(\left(a+b\right).\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\) 4 với a,b > 0
\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\) 9 với a,b,c > 0
\(a^2+b^2+c^2\ge ab+bc+ca\)
1. Chứng minh các bất đẳng thức sau:
a. \(a^2+b^2+c^2\ge ab+bc+ca\)
b. \(a^2+b^2+c^2+d^2\ge ab+bc+cd+da\)
c. \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
2. Cho x,y,z không âm. Cmr: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)
3. Cho a+b+c=1. Cm: \(a^2+b^2+c^2\ge\dfrac{1}{3}\)
cho a,b\(\ge\)1 chứng minh\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{1}{1+ab}\)
CMR:
a,(\(a^4+b^4\)) ≥ \(\left(a+b\right)^4\)
b,\(\left(a^2+b^2\right)\)≥ \(ab\left(a+b\right)^2\)
c, \(a^2+b^2+c^2\)≥ a(b+c)
d, \(a^2+b^2+c^2+d^2\)≥ a(b+c+d)