Another question!!!?

Question

If a, b, c are 3 positive real numbers, prove

(a^2 + 1)/(b + c) + (b^2 + 1)/(c + a) + (c ^2 + 1)/(a + b) >or = 3

And please answer my other questions.

Answer 1

OK, here's the proof ( I HATE THIS QUESTION)....

First, let A=b+c, B=a+c, and C=a+b

We will show the stronger inequality 2a/A + 2b/B + 2c/C >=3, since

x^2+1>=2x for all x. (This is proved by noting x^2-2x+1>=0 for all since (x-1)^2 is a non-negative number.)

Step 1:

Proof that x^2 / f + y^2 / g >= (x+y)^2 / (f+g)

Multiply through by the denominators, so fx^2(f+g)+gy^2(f+g) >= (x+y)^2 *fg, which will simplify down to (xf-gy)^2 >=0, which is true.

Step 2:

I will omit the proof, but the above case extends to three (and more) terms, so x^2 / f + y^2 / g + z^2 / h >= (x+y+z)^2 / (f+g+h)

To prove it, simply use step 1 twice.

Step 3:

We are now ready to prove the big result.

Remember,A=b+c, B=a+c, and C=a+b
Again, note that

* a^2/A + b^2/B + c^2/C >= 2a/A + 2b/B + 2c/C.

Now, This is equal to 2*[ a^2/aA + b^2/bB + c^2/cC ]. This is, by steps 1 and 2 above,

>= 2(a+b+c)^2 / (aA+bB+cC).

Note aA = ab+ac, bB = ab+bc, cC = ac+bc, so their sum = 2(ab+ac+bc), so we need to show, after all this, that (after cancelling 2's)

** (a+b+c)^2 / (ab+ac+bc) >= 3.

Expanding the square, and moving the denominator to the right, we get

a^2 + b^2 + c^2 + 2(ab + ac + bc) >= 3(ab + ac + bc)

So a^2 + b^2 + c^2 - ab - ac - bc >= 0.

But this is exactly[ (a-b)^2 + (a-c)^2 + (b-c)^2 ] / 2, which is always >=0, so the damn thing is proved.

Steve

Answer 2

yet another Brick contained in the Wall (parts a million, 2, 3) - purple Floyd and an awesome conceal with the aid of Korn one way Or yet another - Blondie yet another One Bites The dirt - Queen thoughts of yet another damaged living house - eco-friendly day (section 5 of the Jesus of Suburbia medley) no longer yet another Teenage Anthem - Wednesday 13

Answer 3

what a chore!