SOLUTION: (a+b+c+d)^2<=4(a^2+b^2+c^2+d^2)

Question 743374: (a+b+c+d)^2<=4(a^2+b^2+c^2+d^2)
Found 2 solutions by Menjax, mathyguy:
Answer by Menjax(62) (Show Source): You can put this solution on YOUR website!
What is the question?
Answer by mathyguy(1) (Show Source): You can put this solution on YOUR website!
Omg, he is clearly asking you to prove it, duh. You are clearly not an Algebra master if you don't even know this simple identity, I think you are some kind of troller.
By the Cauchy-Schwarz Inequality we have:
(a+b+c+d)^2<=(1+1+1+1)(a^2+b^2+c^2+d^2)
RELATED QUESTIONS
If a=2,b=3,c=4,d=5:solve... (answered by ValorousDawn)
If (a+b)^2 + (b+c)^2 + (c+d)^2 = 4(ab+bc+cd), prove... (answered by mananth)
1. (A->B) v (C�D) 2.A... (answered by jim_thompson5910)
1. A -> (B -> C) 2. A -> B 3. ~ C ->(A V D) / C V... (answered by Edwin McCravy)
Let a*b=(-1)^ab Let c#d=(c)^d^2 Compute... (answered by Alan3354)
which solution is positive a) -2^2 b) (-2)^3 c)(-2)^4... (answered by Alan3354)
A= 2, b = 3, c = 4, d = 5 3ab + 2cd... (answered by algebrahouse.com)
(a+b)^2-(c+d)^2 How do you... (answered by flame8855)
Solve for (a,b,c,d) a+b+c=0 b+c+d=1 a+c+d=2... (answered by ankor@dixie-net.com)